PEARSON
PHYSICS
NEW SOUTH WALES
STUDENT BOOK
NSW
STAGE 6
i
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PHYSICS
NEW SOUTH
SOUTH WALES
WALES
STUDENT BOOK
Writing and developmentPHYSICS
team
PEARSON
NEW SOUTH WALES
STUDENT BOOK
We are grateful to the following people for their time and expertise in
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PEARSON PHYSICS 11 NEW SOUTH WALES STUDENT BOOK
PEARSON PHYSICS 11 NEW SOUTH WALES STUDENT BOOK
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NEW SOUTH WALES
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STAGE 6
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STAGE 6
Teacher
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iii
Working scientifically
CHAPTER 1 Working scientifically
Module 2 Dynamics
2
CHAPTER 4 Forces
119
1.1
Questioning and predicting
1.2
Planning investigations
10
How are forces produced between objects
and what effects do forces produce?
1.3
Conducting investigations
15
4.1
Newton’s first law
120
1.4
Processing data and information
19
4.2
Newton’s second law
129
1.5
Analysing data and information
24
4.3
Newton’s third law
135
1.6
Problem solving
33
Chapter 4 review
141
1.7
Communicating
36
Chapter 1 review
42
4
Module 1 Kinematics
CHAPTER 2 Motion in a straight line
47
How is the motion of an object moving in a
straight line described and predicted?
CHAPTER 5 Forces, acceleration and energy
143
How can the motion of objects be explained
and analysed?
5.1
Forces and friction
144
5.2
Work
155
5.3
Energy changes
162
5.4
Mechanical energy and power
170
Chapter 5 review
181
2.1
Scalars and vectors
48
2.2
Displacement, speed and velocity
55
2.3
Acceleration
63
2.4
Graphing position, velocity and acceleration
over time
68
How is the motion of objects in a simple system
dependent on the interaction between the objects?
2.5
Equations of motion
78
6.1
Conservation of momentum
184
2.6
Vertical motion
84
6.2
Change in momentum
193
Chapter 2 review
89
6.3
Momentum and net force
196
Chapter 6 review
204
CHAPTER 3 Motion on a plane
93
Module 2 review
How is the motion of an object that changes its
direction of movement on a plane described?
3.1
Vectors in two dimensions
3.2
Vector components
103
3.3
Relative motion
106
Chapter 3 review
111
Module 1 review
iv
CHAPTER 6 Momentum, energy and
simple systems
94
113
183
206
Module 3 Waves and
thermodynamics
CHAPTER 7 Wave properties
Module 4 Electricity and
magnetism
215
What are the properties of all waves and wave motion?
CHAPTER 12 Electrostatics
337
How do charged objects interact with other
charged objects and with neutral objects?
7.1
Mechanical waves
216
7.2
Measuring mechanical waves
220
12.1 Electric charge
338
Chapter 7 review
227
12.2 Electric fields
344
12.3 Coulomb’s law
352
CHAPTER 8 Wave behaviour
229
How do waves behave?
8.1
Wave interactions
230
8.2
Resonance
240
Chapter 8 review
243
CHAPTER 9 Sound waves
245
What evidence suggests that sound is a
mechanical wave?
9.1
Sound as a wave
246
9.2
Sound behaviour
252
9.3
Standing waves
259
Chapter 9 review
268
CHAPTER 10 Ray model of light
271
What properties can be demonstrated when
using the ray model of light?
CHAPTER 13 Electric circuits
272
10.2 Refraction
278
10.3 Curved mirrors and lenses
284
Chapter 10 review
296
299
357
359
How do the processes of the transfer and the
transformation of energy occur in electric circuits?
13.1 Electric current and circuits
360
13.2 Energy in electric circuits
367
13.3 Resistance
374
13.4 Series and parallel circuits
384
Chapter 13 review
397
CHAPTER 14 Magnetism
399
How do magnetised and magnetic objects interact?
14.1 Magnetic materials
400
14.2 Magnetic fields
404
14.3 Calculating magnetic fields
10.1 Light as a ray
CHAPTER 11 Thermodynamics
Chapter 12 review
Chapter 14 review
Module 4 review
413
417
419
ANSWERS
423
GLOSSARY
437
INDEX
441
How are temperature, thermal energy and
particle motion related?
11.1 Heat and temperature
300
11.2 Specific heat capacity
308
11.3 Latent heat
312
11.4 Conduction
317
11.5 Convection
322
11.6 Radiation
Chapter 11 review
Module 3 review
325
328
330
v
How to use this book
Pearson Physics 11
New South Wales
CHAPTER
Forces
In the seventeenth century Sir Isaac Newton published three laws that explain why
objects in our universe move as they do. These laws became the foundation of a
branch of physics called mechanics: the science of how and why objects move. They
have become commonly known as Newton’s three laws of motion.
Pearson Physics 11 New South Wales
has been written to the new New
South Wales Physics Stage 6 Syllabus.
The book covers Modules 1 to 4 in
an easy-to-use resource. Explore
how to use this book below.
Using Newton’s laws, this chapter will describe the relationship between the forces
acting on an object and its motion.
Content
INQUIRY QUESTION
How are forces produced between objects and what effects do
forces produce?
By the end of this chapter you will be able to:
Section
• using Newton’s Laws of Motion, describe static and dynamic interactions between
two or more objects and the changes that result from:
- a contact force
- a force mediated by fields
• explore the concept of net force and equilibrium in one-dimensional and simple
two-dimensional contexts using: (ACSPH050) ICT N
- algebraic addition
- vector addition
- vector addition by resolution into components
• solve problems or make quantitative predictions about resultant and component
forces by applying the following relationships: ICT N
- F AB = −F BA
- F x = F cos θ , F y = F sinθ
• conduct a practical investigation to explain and predict the motion of objects on
inclined planes (ACSPH098) CCT ICT
Chapter opener
Each chapter is clearly divided
into manageable sections of
work. Best-practice literacy and
instructional design are combined
with high quality, relevant photos
and illustrations to help students
better understand the idea or
concept being developed.
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
The chapter opening page link
the Syllabus to the chapter
content. Key content addressed
in the chapter is clearly listed.
M04_PPN_SB11_9298.indd 119
PHYSICS INQUIRY
CCT
Ray model of light
What properties can
be demonstrated when
using the ray model of
light?
11/7/17 12:03 PM
10.2 Refraction
In musical instruments and loudspeakers, resonance is a desired effect. The
sounding boards of pianos and the enclosures of loudspeakers are designed to
enhance and amplify particular frequencies. In other systems, such as car exhaust
systems and suspension bridges, resonance is not always desirable, and care is taken
to design a system that prevents resonance.
When light passes from one medium (substance) into another, it either speeds up or
slows down. This change in speed causes the light ray to change direction, as shown
in Figure 10.2.1. Refraction is the name given to a change in the direction of light
caused by changes in its speed (Figure 10.2.1).
PHYSICSFILE ICT CCT
Tacoma Narrows Gorge suspension bridge
One of the most recognisable cases of mechanical resonance for many years has been
the destruction of the Tacoma Narrows Gorge Bridge in the US State of Washington in
1940. It was originally
thought, from studying
video footage of the
bridge’s collapse that the
wind acted as a driving
frequency, causing the
bridge to oscillate with
ever-increasing amplitude
until the whole bridge
shook itself apart. More
recent research seems to
suggest that instead the
wind supplied a twisting
motion causing the bridge FIGURE 8.2.3 The Tacoma Narrows Gorge Bridge, Washington,
U.S.A. bends and buckles due to the force of wind.
to tear itself apart.
COLLECT THIS…
•
a large beaker
•
laser light
•
clear water beads
•
white paper
DO THIS…
1 Soak the water beads in water.
2 Place the beaker on the white
paper and half fill with water.
3 Shine a laser light through the
water at an angle. On the paper,
mark the path into and out of
the beaker.
4 Place the water beads into the
beaker. Adjust the water so that
half of the beads are covered
in water and half are out of
the water.
5 Using the laser pointer, shine
through the bottom of the
beaker at the same angle
initially marked on the paper.
Mark the path of the light on
the paper in a different colour.
6 Using the laser pointer, shine
through the top of the beaker.
How does the path the light
takes differ?
FIGURE 10.2.1 Light refracts as it moves from one medium (i.e. the semicircular glass prism) into
another (i.e. air) causing a change in direction.
REFRACTIVE INDEX
TABLE 10.2.1
Speed of light (× 108 m s−1)
vacuum
3.00
air
3.00
ice
2.29
Present your results describing the
path of the light.
water
2.25
quartz
2.05
REFLECT ON THIS…
crown glass
1.97
What properties can be
demonstrated when using the ray
model of light?
flint glass
1.85
diamond
1.24
Describe how we see clear objects.
WE
ICT
Resonance in aircraft wings
Have you ever looked out of an aeroplane window and
noticed that the wings are vibrating up and down? This
effect, sometimes known as flutter, is due to the vibrational
energy both of the engines on the aeroplane and the
air flow across the wings. While small vibrations in the
wings are normal, resonance in the wings is not a desired
effect. Every effort is made to make sure that the driving
frequency of the engines and the driving frequency of
the air flow do not match the natural resonant frequency
of the wings. Aeronautical engineers do not want the
energy from the driving vibrations to be transferred to the
aeroplane wings. Engineers and pilots test aeroplanes by
flying them at great speeds, often close to the speed of
sound, to see how they manage the vibrations that such air
flow causes.
The speed of light in various materials correct to three significant figures.
Material
RECORD THIS…
PHYSICS IN ACTION
The amount of refraction that occurs depends on how much the speed of light
changes as light moves from one medium to another—when light slows down
greatly, it will undergo significant refraction.
The speed of light in a number of different materials is shown in Table 10.2.1.
FIGURE 8.2.4 Small vibrations in aeroplane wings are expected.
Resonance in aeroplanes is not a desired effect.
When you look at a glass, how can
you see if the glass is full of air,
water or another clear substance?
278
vi
CHAPTER 8 | WAVE BEHAVIOUR
MODULE 3 | WAVES AND THERMODYNAMICS
M10_PPN_SB11_9298.indd 278
FIGURE 8.2.2 The sound box of a stringed
instrument is tuned to resonate for the range of
frequencies of the vibrations being produced by
the strings. When a string is plucked or bowed,
the airspace inside the box vibrates in resonance
with the natural frequency and the sound is
amplified.
11/7/17 12:12 PM
M08_PPN_SB11_9298.indd 241
241
11/7/17 8:32 AM
Physics Inquiry
Physics in Action
Physics Inquiry features are inquirybased activities that pre-empt
the theory and allow students to
engage with the concepts through
a simple activity that sets students
up to ‘discover’ the science before
they learn about it. They encourage
students to think about what happens
in the world and how science can
provide explanations.
Physics in Action boxes place physics in an applied situation
or a relevant context. These refer to the nature and practice
of physics, applications of physics and the associated issues
and the historical development of concepts and ideas.
PhysicsFile
PhysicsFiles include a range of interesting and
real-world examples to engage students.
Highlight box
Aeroplane in a cross wind
A similar situation can be applied to calculating the velocity of an aeroplane with
respect to the ground. As an aeroplane flies it will experience winds blowing
opposite to its direction of motion (head wind), in the same direction to its motion
(tail wind), or at some angle across its direction of motion (cross wind). If you
know both the velocity of the plane relative to the wind and the velocity of the wind
relative to the ground, by using the rules for vector addition the resultant vector of
these two values will describe the velocity of the plane relative to the ground.
Highlight boxes focus students’
attention on important information
such as key definitions, formulae
and summary points.



vPG = vPW + v WG

where vPG is the velocity of the plane relative to the ground

vPW is the velocity of the plane relative to the wind

v WG is the velocity of the wind relative to the ground
Worked example 3.3.3
FIND THE RESULTANT VELOCITY OF AN AEROPLANE IN A CROSS WIND
A light aircraft is travelling at 300 km h–1 north, with a crosswind blowing at
45.0 km h–1 west.
WS
1.7
Determine the velocity of the plane relative to the ground.
Worked examples
Thinking
Working
Write out the equation describing the
resultant velocity.



vPG = vPW + v WG
Construct a vector diagram showing
the vectors drawn head to tail. Draw
the resultant vector from the tail of
the first vector to the head of the last
vector.
Worked examples are set out in steps that show thinking and
working. This format greatly enhances student understanding
by clearly linking underlying logic to the relevant calculations.
As the two vectors to be added are at
90° to each other, apply Pythagoras’
theorem to calculate the magnitude of
the resultant velocity.
Using trigonometry, calculate the angle
from the west vector to the resultant
vector.
Each Worked example is followed by a Try Yourself activity.
This mirror problem allows students to immediately test their
understanding.
→
WG
ν
N
W
PHYSICSFILE ICT N
Components of flight
= 45 km h–1
The velocity vector that describes the
direction and speed of an aeroplane
can be broken down into multiple
vector components. These are known
as thrust, lift, drag and weight forces.
The thrust is generated by the engines
and gives the plane its forward
motion. The weight force describes the
downwards pull due to gravity. The drag
component slows the plane down as
it pushes through the air. And the lift
component is produced by the wings
and makes the plane rise.
When designing a plane, all of these
components need to be considered.
E
S
→
PG
ν
→
PW
ν
= 300 km h–1
2
= 3002 + 452
v PG
= 90 000 + 2025
vPG = 92 025
= 303 km h–1
tanθ =
WS
1.9
45
300
θ = tan−1 0.15
= 8.53°
Determine the direction of the vector
relative to north or south.
The direction is N8.53°W
State the magnitude and direction of
the resultant vector.

vPG = 303 km h–1, N8.53°W
Worked example: Try yourself 3.3.3
FIND THE RESULTANT VELOCITY OF AN AEROPLANE IN A CROSS WIND
Fully worked solutions to all Worked example: Try yourself are
available on Pearson Physics 11 New South Wales Reader+.
FIGURE 3.3.4 There are multiple vector
components involved in the direction and
speed of an aeroplane.
A jet aircraft is travelling at 900 km h–1 east, with a crosswind blowing at
85.0 km h–1 south.
Determine the velocity of the plane relative to the ground.
CHAPTER 3 | MOTION ON A PLANE
M03_PPN_SB11_9298.indd 109
109
11/7/17 12:00 PM
Additional content
Additional content features include
material that goes beyond the core content
of the Syllabus. They are intended for
students who wish to expand their depth of
understanding in a particular area.
Section summary
Each section has a section summary
to help students consolidate the key
points and concepts of each section.
PHYSICSFILE CCT
Substitute the values for this situation
into the equation.
∆U = 50 × 9.8 × 0.4
Newton’s universal law of
gravitation
State the answer with appropriate
units and significant figures.
∆U = 196 J
Gravitational field strength (N/kg)
The formula ∆U = mg ∆h is based
on the assumption that the Earth’s
gravitational field is constant.
However, Newton’s universal law of
gravitation predicts that the Earth’s
gravitational field decreases with
altitude (Figure 5.3.5). This decrease
only becomes significant far above the
Earth’s surface. Close to the surface the
assumption of a constant gravitational
field is valid.
12
1.1 Review
SUMMARY
• Before you begin your research it is important to
conduct a literature review. By utilising data from
primary and/or secondary sources, you will better
understand the context of your investigation to
create an informed inquiry question.
Worked example: Try yourself 5.3.5
CALCULATING GRAVITATIONAL POTENTIAL ENERGY RELATIVE TO A
REFERENCE LEVEL
A father picks up his baby from its bed. The baby has a mass of 6.0 kg and the
mattress of the bed is 70 cm above the ground. When the father holds the baby in
his arms, it is 125 cm off the ground.
Use g as 9.8 N kg−1 and give your answer correct to two significant figures.
4
2
5 10 15 20 25 30 35
Altitude (× 1000 km)
FIGURE 5.3.5 The
Earth’s gravitational field
strength decreases with altitude.
SKILLBUILDER
N
Multiplying
vectors
When two vectors are multiplied
together, the result is a scalar
variable. For instance, in the

equation for work, W = Fnet s , the
force vector is multiplied by the
displacement vector to produce
the scalar quantity of work.
- A dependent variable is a variable that
may change in response to a change in the
independent variable. This is the variable that
will be measured or observed.
• Once a question has been chosen, stop to evaluate
the question before progressing. The question
may need further refinement or even further
investigation before it is suitable as a basis for
an achievable and worthwhile investigation.
Elastic potential energy
6
- An independent variable is a variable that is
selected by the researcher and changed during
the investigation.
• The hypothesis is a testable prediction based on
previous knowledge and evidence or observations,
and attempts to answer the inquiry question.
+ ADDITIONAL
8
• There are three categories of variables:
• The purpose is a statement describing what
is going to be investigated. For example: ‘The
purpose of the experiment is to investigate the
relationship between force, mass and acceleration.’
Calculate the increase in gravitational potential energy of the baby.
Earth's gravitational field
strength with altitude
10
It is important not to attempt something that you
cannot complete in the time available or with
the resources on hand. For example, it might be
difficult to create a complicated device with the
facilities available in the school laboratory.
Another important form of potential energy is elastic potential energy. Elastic
potential energy can be stored in many ways; for example, when a spring is
stretched, a rubber ball is squeezed, air is compressed in a tyre, or a bungee
rope is extended during a jump.
- A controlled variable is a variable that is kept
constant during the investigation.
• It is important to change only one independent
variable during the investigation.
KEY QUESTIONS
Materials that have the ability to store elastic potential energy when work is
done on them, and then release this energy, are called elastic materials. Metal
springs and bouncing balls are common examples; however, many other
materials are at least partially elastic. If their shape is manipulated, items such
as our skin, metal hair clips and wooden rulers all have the ability to restore
themselves to their original shape once released.
Materials that do not return to their original shape and release their stored
potential energy are referred to as plastic materials. Plasticine is an example
of a very plastic material.
1
Scientists make observations from which a hypothesis
is stated and this is then experimentally tested.
a Define ‘hypothesis’.
b How are theories and principles different from a
hypothesis?
2
Which of the following describes an inquiry question?
A If an object is subject to a constant net force, then it
will move with a constant acceleration.
B What features suggest that sound is a mechanical
wave?
C Increasing the voltage in an electric circuit causes
an increase in the current.
D The momentum in an inelastic collision was
conserved.
3
In a practical investigation, a student changes the
voltage by adding or subtracting batteries in series to
the circuit.
a How could the voltage be a discrete variable?
b How could it be a continuous variable?
The elastic potential energy of an object, Ep, is given by the formula:
1 2
kx
2
Ep =
where:
k is a property of the elastic material called the spring constant
x is the amount of extension or compression of the material
Similarly, in the equation for
1 
kinetic energy, K = mv 2, the two
2
4
In another experiment a student uses the following
range of values to describe the brightness of a light:
dazzling, bright, glowing, dim, off
What type of variable is ‘brightness’?
5
Select the best hypothesis from the three options
below. Give reasons for your choice.
A Hypothesis 1: If both the angular momentum and
inertia of a rotating system are increased, then the
angular (rotational) velocity will also increase.
B Hypothesis 2: Your position during angular airborne
motion affects your inertia.
C Hypothesis 3: If rotational velocity increases as
radius decreases, then a springboard diver’s
angular (rotational) velocity is slower when they
hold a stretched (layout) position than when they
are in a tuck position, if they take off with the same
angular momentum.
velocity vectors are squared to
create the scalar value for energy.
26
MODULE 2 | DYNAMICS
M05_PPN_SB11_9298.indd 26
CHAPTER 1 | WORKING SCIENTIFICALLY
11/1/17 6:10 PM
M01_PPN_SB11_9298.indd 9
9
11/7/17 11:50 AM
SkillBuilder
Section review questions
A skillBuilder outlines a method or
technique. They are instructive and selfcontained. They step students through the
skill to support science application.
Each section finishes with key questions
to test students’ understanding and ability
to recall the key concepts of the section.
vii
How to use this book
Module review
Each module finishes with a set of questions,
including multiple choice, short answer and
extended response. These questions assist
students in drawing together their knowledge
and understanding, and applying it to these
types of questions.
Chapter review
Each chapter finishes with a list of key terms
covered in the chapter and a set of questions
to test students’ ability to apply the knowledge
gained from the chapter.
MODULE 1 • REVIEW
Chapter review
REVIEW QUESTIONS
mass
net force
newton
Newton’s first law
Newton’s second law
contact force
equilibrium
force
force mediated by a field
inertia
Newton’s third law
normal reaction force
terminal velocity
weight
Multiple choice
1
A car accelerates in a straight line at a rate of 5.5 m s
from rest. What distance has the car travelled at the end
of three seconds?
A 8.25 m
B 11 m
C 16.5 m
D 24.75 m
7
Consider three forces acting on a single object: a 20 N
upwards force, a 10 N downwards force, and a 10 N
force from left to right. Sketch:
a the vector diagram of the three forces and the
resultant force
b the force required for the object to be in
equilibrium.
2
A bike accelerates in a straight line at a rate of 2.5 m s−2
from rest. What distance does the bike travel in the third
second of its motion?
A 6.25 m
B 13.75 m
C 19.25 m
D 24.75 m
8
If two equal masses experience the same force, which
of the following describes their accelerations?
A equal and opposite
B equal and in the same direction
C different and opposite
D different and in the same direction
3
A graph depicting the velocity of a small toy train versus
time is shown below. The train is moving on a straight
section of track, and is initially moving in an easterly
direction.
KEY QUESTIONS
Which of the following are examples of contact forces?
A two billiard balls colliding
B the electrostatic force between two charged
particles
C a bus colliding with a car
D the magnetic force between two fridge magnets
2
A student is travelling to school on a train. When
the train starts moving, she notices that passengers
tend to lurch towards the back of the train before
regaining their balance. Has a force acted to push the
passengers backwards? Explain your answer.
A bowling ball rolls along a smooth wooden floor at
constant velocity. Which of the following diagrams
correctly indicates the horizontal forces acting on
the ball as it rolls towards the right? (The weight and
normal force can been ignored.)
A
C
4
5
6
B
F⃗
F⃗
F⃗
D
F⃗
9
Calculate the mass of an object if it accelerates at
9.20 m s−2 east when a force of 352 N east acts on it.
F⃗
F⃗
Newton’s first law states that an object will maintain a
constant velocity unless an unbalanced, external force
acts on it. What distinguishes an external force from an
internal force?
12 A 150 N force acts at a 45° angle to the x direction on
an object with a mass of 10 kg. A second force of 15 N
acts on the same object at an angle of 30° to the x
direction. Using the diagram below, calculate the net
force and initial acceleration acting on the object in the
x direction.
→
F2
30º
y
x
→
F1
45º
What are the horizontal and vertical components
of a force of 50 N acting on an object at an angle of
45° upwards from the positive x direction?
CHAPTER 4 | FORCES
M04_PPN_SB11_9298.indd 141
0
2
4
A ball is dropped, falls vertically and strikes the ground
with a velocity of +5 m s−1. It rebounds, and leaves the
ground with a velocity of −3 m s−1. What is the change in
velocity that the ball experiences?
A −8 m s−1
B +8 m s−1
C −2 m s−1
D +2 m s−1
6
An aeroplane flies a distance of 300 km due north, then
changes course and travels 400 km due east.
What is the distance travelled and the final displacement
of the aeroplane?
A distance = 700 km, displacement = 500 km north-east
B distance = 700 km, displacement = 700 km north-east
C distance = 700 km, displacement = 500 km N53.1°E
D distance = 700 km, displacement = 500 km N36.9°E
8
10
12
7
A car that is initially at rest begins to roll down a steep
road that makes an angle of 11° with the horizontal.
Assuming a constant acceleration of 2 m s−1, what is the
speed of the cr after it has travelled 100 metres?
A 19 m s−1
B 20 km h−1
C 72 km h−1
D 72 m s−1
8
Which equation can be used to calculate the velocity of
a boat relative to a submarine? (Use the subscripts B for
boat, S for submarine and G for ground.)



A vBS = vBG + v SG



B vBS = vSG + vBS



C vBS = vSG + ( −vBG )



D vBS = vBG + ( −v SG )
Time (s)
–0.1
–0.2
a What distance does the train travel in the first
6 seconds of its motion?
A 0m
B 0.4 m
C 0.8 m
D 1.2 m
b What is the displacement of the train after the first
11 seconds of its motion?
A 0m
B 0.4 m east
C 0.8 m east
D 1.2 m east
141
11/7/17 12:04 PM
6
WS
1.11
B 2v
C 2v
D 4v
+0.1
10 Calculate the acceleration of a 60.9 g golf ball when a
net force of 95.0 N south acts on it.
WS
1.10
A ball dropped from rest from a height h hits the
ground with a speed v. The ball is then released from a
height of 2h. With what speed would the ball now strike
the ground?
A 12 v
5
+0.2
11 Calculate the acceleration of a 657 kg motorbike when
a net force of 3550 N north acts on it.
A force of 10 N acts from left to right on an object, and
a force of 5 N simultaneously acts from right to left on
the same object.
a What is the net force acting on the object?
b Are the forces in equilibrium?
4
−2
Velocity of train (m s–1)
1
3
MR
1
Kinematics
KEY TERMS
REVIEW QUESTIONS
M03A_PPN_SB11_9298.indd 113
113
11/7/17 9:45 AM
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Digital
1
CHAPTER
CHAPTER
Working scientifically
This chapter covers the skills needed to successfully plan and conduct investigations
using primary and secondary sources.
1.1 Questioning and predicting explains how to develop, propose and evaluate
inquiry questions and hypotheses. When creating a hypothesis, a consideration of
the variables must be included.
1.2 Planning investigations explores how to identify risks and to make sure all
ethical concerns are considered. It is important to choose appropriate materials and
technology to carry out your investigation. You will also need to confirm that your
choice of variables allows for a reliable collection of data.
1.3 Conducting investigations describes methods for accurately collecting and
recording data to reduce errors. Appropriate procedures need to be carried out
when disposing of waste.
1.4 Processing data and information describes ways to present your data from
an array of visual representations. You will learn to identify trends and patterns in
your data.
1.5 Analysing data and information explains error and uncertainty and how to
construct mathematical models to better understand the scientific principles of your
research.
1.6 Problem solving describes how to understand the scientific principles underlying
the solution to your inquiry question.
1.7 Communicating explains how to appropriately use scientific language,
nomenclature and scientific notation and describes different forms of
communication.
Outcomes
By the end of this chapter you will be able to:
• develop and evaluate questions and hypotheses for scientific investigation
(PH11-1)
• design and evaluate investigations in order to obtain primary and secondary data
and information (PH11-2)
• conduct investigations to collect valid and reliable primary and secondary data
and information (PH11-3)
• select and process appropriate qualitative and quantitative data and information
using a range of appropriate media (PH11-4)
• analyse and evaluate primary and secondary data and information (PH11-5)
• solve scientific problems using primary and secondary data, critical thinking skills
and scientific processes (PH11-6)
• communicate scientific understanding using suitable language and terminology
for a specific audience or purpose (PH11-7).
Content
By the end of this chapter you will be able to:
• develop and evaluate inquiry questions and hypotheses to identify a concept
that can be investigated scientifically, involving primary and secondary data
(ACSPH001, ACSPH061, ACSPH096) L
• modify questions and hypotheses to reflect new evidence CCT
• assess risks, consider ethical issues and select appropriate materials and
technologies when designing and planning an investigation (ACSPH031,
ACSPH097) EU PSC
• justify and evaluate the use of variables and experimental controls to ensure
that a valid procedure is developed that allows for the reliable collection of data
(ACSPH002)
• evaluate and modify an investigation in response to new evidence CCT
• employ and evaluate safe work practices and manage risks
(ACSPH031) PSC WE
• use appropriate technologies to ensure and evaluate accuracy ICT N
• select and extract information from a wide range of reliable secondary sources
and acknowledge them using an accepted referencing style L
• select qualitative and quantitative data and information and represent them
using a range of formats, digital technologies and appropriate media (ACSPH004,
ACSPH007, ACSPH064, ACSPH101) L N
• apply quantitative processes where appropriate N
• evaluate and improve the quality of data CCT N
• derive trends, patterns and relationships in data and information
• assess error, uncertainty and limitations in data (ACSPH004, ACSPH005,
ACSPH033, ACSPH099) CCT
• assess the relevance, accuracy, validity and reliability of primary and secondary
data and suggest improvements to investigations (ACSPH005) CCT N
• use modelling (including mathematical examples) to explain phenomena, make
predictions and solve problems using evidence from primary and secondary
sources (ACSPH006, ACSPH010) CCT
• use scientific evidence and critical thinking skills to solve problems CCT
• select and use suitable forms of digital, visual, written and/or oral forms of
communication L N
• select and apply appropriate scientific notations, nomenclature and scientific
language to communicate in a variety of contexts (ACSPH008, ACSPH036,
ACSPH067, ACSPH102) L N
• construct evidence-based arguments and engage in peer feedback to evaluate an
argument or conclusion (ACSPH034, ACSPH036). CC DD
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
1.1 Questioning and predicting
Before you are able to start the practical side of your investigation, you first need
to understand the theory behind it. This section is a guide to some of the key steps
that should be taken when first developing your inquiry questions and hypotheses.
DEVELOPING AN INQUIRY QUESTION AND PURPOSE,
FORMULATING HYPOTHESES AND MAKING PREDICTIONS
The inquiry question, purpose and hypothesis are interlinked. It is important to
note that each of these can be refined as the planning of the investigation continues.
Inquiry questions, purposes and hypotheses
The inquiry question defines what is being investigated. For example:
‘What is the relationship between voltage and current in a DC circuit?’
The purpose is a statement describing what is going to be investigated. For
example:
‘The purpose of the experiment is to investigate the relationship between voltage
and the current in a circuit of constant resistance.’
The hypothesis is a testable prediction based on previous knowledge and
evidence or observations, and attempts to answer the inquiry question. For example:
‘If voltage is directly proportional to current in a circuit of constant resistance
and you increase the voltage, then the current will also increase.’
Formulating a question
Before formulating a question, it is good practice to conduct a literature review of
the topic to be investigated. You should become familiar with the relevant scientific
concepts and key terms.
During this review, write down questions or correlations as they arise.
Compile a list of possible ideas. Do not reject ideas that initially might seem
impossible. Use these ideas to generate questions that are answerable.
Before constructing a hypothesis, decide on a question that needs an answer.
This question will lead to a hypothesis when:
• the question is reduced to measurable variables
• a prediction is made based on knowledge and experience.
The different types of variables are discussed on pages 6–7.
FIGURE 1.1.1 There are many elements to
a practical investigation, which may appear
overwhelming to begin with. Taking a step-bystep approach will help the process and assist in
completing a solid and worthwhile investigation.
4
Evaluating your question
Once a question has been chosen, stop to evaluate the question before progressing.
The question may need further refinement or even further investigation before it
is suitable as a basis for an achievable and worthwhile investigation. It is important
not to attempt something that you cannot complete in the time available or with the
resources on hand. For example, it might be difficult to create a complicated device
with the facilities available in the school laboratory.
To evaluate the question, consider the following:
• Relevance: Is the question related to the area of study?
• Clarity and measurability: Can the question be framed as a clear hypothesis?
If the question cannot be stated as a specific hypothesis, then it is going to be
very difficult to complete the research.
• Time frame: Can the question be answered within a reasonable period of time?
Is the question too broad?
• Knowledge and skills: Do you have the knowledge and skills that will allow you
to answer the question? Keep the question simple and achievable.
CHAPTER 1 | WORKING SCIENTIFICALLY
•
•
•
Practicality: Are the resources such as laboratory equipment and materials you
will need likely to be readily available? Keep things simple. Avoid investigations
that require sophisticated or rare equipment. Readily available equipment
includes timing devices, objects that could be used as projectiles, a tape measure
and other common laboratory equipment.
Safety and ethics: Consider the safety and ethical issues associated with the
question you will be investigating. If there are issues, can these be addressed?
Advice: Seek advice from the teacher about the question. Their input may prove
very useful. Their experience may lead them to consider aspects of the question
that you have not thought about.
Sourcing information
Once you have selected a topic, the next step is to source reliable information. Some
of the steps involved in sourcing information are:
• identifying key terms
• evaluating the credibility of sources
• evaluating experimental data/evidence.
Sources can be:
• primary sources—original sources of data and evidence generated by a person
or group directly; for example, by personally conducting a practical investigation
• secondary sources—analyses and interpretations of primary sources; for
example, textbooks, magazine articles and newspaper articles. This also includes
interpreting other people’s experimental data such as reports, graphs and
diagrams.
Some of the sources that may contain useful information include:
• newspaper articles and opinion pieces
• journal articles (from peer-reviewed journals)
• magazine articles
• government reports
• global databases, statistics and surveys
• laboratory work
• computer simulations and modelling
• interviews with professionals (e.g. on-line or by email)
Some reputable science journals and magazines are:
• Cosmos
• Double Helix
• ECOS
• Nature
• New Scientist
• Popular Science
• Scientific American.
Hypothesis
A hypothesis is a prediction, based on evidence and prior knowledge or
observations, that attempts to answer the inquiry question. A hypothesis often takes
the form of a proposed relationship between two or more variables in a cause-andeffect relationship, such as ‘If X happens, then Y will happen.’
Here
aresome examples of hypotheses:
• If F = ma , then for a constant force, when mass is increased the acceleration of
an object will decrease.
• Assuming that all objects fall at the same speed due to gravity, if two objects are
simultaneously dropped from the same height, they will both land at the same
time.
Hypotheses can be written
in a variety of ways, such as ‘A
happens because of B’, or ‘when A
happens, B will happen’. However
they are written, hypotheses
must always be testable and must
clearly state the independent and
dependent variables.
CHAPTER 1 | WORKING SCIENTIFICALLY
5
•
If velocity increases when radius decreases, then a gymnast who has a set angular
momentum when in the air will rotate faster during a somersault when they tuck
their legs in towards their chest than if they keep their legs stretched out.
It is important to keep in mind that your hypothesis is only a prediction, so
you may find that after conducting your investigation your hypothesis is actually
incorrect. In that case, rather than supporting your hypothesis with your results, you
would analyse your data to explain what you found and re-evaluate the hypothesis.
Variables
A good scientific hypothesis can be tested (that is, supported or refuted) through
investigation. To be a testable hypothesis, it should be possible to measure both what
is changed or carried out and what will happen. The factors that are monitored
during an experiment or investigation are called variables. An experiment or
investigation determines the relationship between variables and measures the results.
There are three categories of variables:
• An independent variable is a variable that is selected by the researcher and
changed during the investigation.
• A dependent variable is a variable that may change in response to a change in
the independent variable. This is the variable that will be measured or observed.
• A controlled variable is a variable that is kept constant during the investigation.
It is important to change only one independent variable during the investigation.
Otherwise you might not be able to tell which variable caused the changes you
observed.
The following is an example of a typical investigation.
Prediction: For a projectile launched into the air at a constant speed, the horizontal
distance it travels will be greatest when the launch angle is 45°.
• independent variable: launch angle
• dependent variable: horizontal distance travelled
• controlled variables: launch speed, mass of projectile, air resistance (including wind)
Completing a table like Table 1.1.1 will assist in evaluating the inquiry question
or questions. In this investigation a marble is launched using a spring-release
mechanism inside a tube.
TABLE 1.1.1
Break the question down to determine the variables.
Inquiry question
How does the angle of release of an arrow affect its projectile
motion?
Hypothesis
If the horizontal distance a projectile reaches is dependent on
the velocity and the launch angle and the initial velocity is kept
constant, a maximum horizontal distance will be reached when
the launch angle of a projectile is 45°.
Independent variable
angle of launch
Dependent variable
horizontal distance travelled
Controlled variables
mass of the arrow
tension in the bow string before launch (i.e. initial velocity of the
arrow)
Qualitative and quantitative variables
Variables are either qualitative or quantitative, with further subsets in each category.
• Qualitative variables (sometimes called categorical variables) can be observed
but not measured. They can only be sorted into groups or categories such as
brightness, type of material or type of device.
• Nominal variables are qualitative variables in which the order is not important;
for example, the type of material or type of device.
• Ordinal variables are qualitative variables in which order is important and groups
have an obvious ranking or level; for example, brightness (Figure 1.1.2).
6
CHAPTER 1 | WORKING SCIENTIFICALLY
FIGURE 1.1.2 When you record qualitative data, describe in detail how each variable will be defined.
For example, if you are recording the brightness of light globes, pictures are a good way of clearly
defining what each assigned term represents.
•
•
•
Quantitative variables can be measured. Length, area, weight, temperature
and cost are all examples of quantitative data.
Discrete variables are quantitative variables that consist only of integer numerical
values (i.e. whole numbers); for example, the number of pins in a packet, the
number of springs connected together, or the energy levels in atoms.
Continuous variables are quantitative variables that can have any numerical value
within a given range; for example, temperature, length, weight, or frequency.
Formulating a hypothesis
Once the inquiry question is confirmed, formulating a hypothesis comes next.
A hypothesis requires a proposed relationship between two variables. It should
predict that a relationship exists or does not exist.
Identify the two variables in your question. State the independent and dependent
variables.
For example: If I do/change this (independent variable), then this (dependent
variable) will happen.
A good hypothesis should:
• be a statement
• be based on information contained in the inquiry question or purpose
• be worded so that it can be tested in the experiment
• include an independent and a dependent variable
• include variables that are measurable.
The hypothesis should also be falsifiable (able to be disproved). This means that
a negative outcome would disprove it. For example, the hypothesis in Table 1.1.1
would be disproved if you found that an angle of 30° resulted in the greatest distance
travelled. Unfalsifiable hypotheses cannot be proved by science. These include
hypotheses on ethical, moral and other subjective judgements.
CHAPTER 1 | WORKING SCIENTIFICALLY
7
Modifying a hypothesis
As you collect new evidence from secondary sources, it may become necessary to
adjust your inquiry question or hypothesis. For example, your hypothesis may be:
‘If objects all accelerate under gravity at the same rate, then objects with different
masses dropped from the same height will land at the same time.’
As you continue your research of secondary sources, you may find that you did not
take into account air resistance when formulating your hypothesis, so you could
modify your hypothesis to:
‘If objects all accelerate under gravity at the same rate, then objects with different
masses and negligible air resistance that are dropped from the same height will land
at the same time.’
Defining the purpose of the investigation
Defining the purpose is a key step in testing the hypothesis. The purpose should
directly relate to the variables in the hypothesis, and describe how each will be
measured. The purpose does not need to include the details of the method.
Example
• Hypothesis 1: If F = ma , then when the force is kept constant, the acceleration
decreases as the mass increases.
Extension: When the force is kept constant, doubling the mass halves the
acceleration.
• Hypothesis 2: When the mass is kept constant, the acceleration increases with
increasing force.
Extension: When the mass is kept constant, doubling the force doubles the
acceleration.
• Purpose: The purpose of the experiment is to investigate the relationship
between force, mass and acceleration.
In the first stage of the experiment, mass will be the independent variable (select
a number of different masses) and the force is constant. The resulting acceleration
(dependent variable) will be measured.
Then in the second stage of the experiment, force will be the independent
variable (you select a number of different forces) and the mass will be kept constant.
The resulting acceleration (dependent variable) will be measured.
These two investigations when combined create the classic Newton’s second law
experiment.
• Hypothesis 1 should give a result that mass is inversely proportional to the
acceleration.
• Hypothesis 2 should give a result that force is proportional to the acceleration.
Using the data collected from both stages of the experiment, the relationship
between the three variables can be determined.
This level of ‘neatness’ is not always possible, especially with a student-designed
experiment, but you should strive towards this.
8
CHAPTER 1 | WORKING SCIENTIFICALLY
1.1 Review
SUMMARY
• Before you begin your research it is important to
conduct a literature review. By utilising data from
primary and/or secondary sources, you will better
understand the context of your investigation to
create an informed inquiry question.
It is important not to attempt something that you
cannot complete in the time available or with
the resources on hand. For example, it might be
difficult to create a complicated device with the
facilities available in the school laboratory.
• The purpose is a statement describing what
is going to be investigated. For example: ‘The
purpose of the experiment is to investigate the
relationship between force, mass and acceleration.’
• There are three categories of variables:
-- An independent variable is a variable that is
selected by the researcher and changed during
the investigation.
• The hypothesis is a testable prediction based on
previous knowledge and evidence or observations,
and attempts to answer the inquiry question.
-- A dependent variable is a variable that
may change in response to a change in the
independent variable. This is the variable that
will be measured or observed.
• Once a question has been chosen, stop to evaluate
the question before progressing. The question
may need further refinement or even further
investigation before it is suitable as a basis for
an achievable and worthwhile investigation.
-- A controlled variable is a variable that is kept
constant during the investigation.
• It is important to change only one independent
variable during the investigation.
KEY QUESTIONS
1
Scientists make observations from which a hypothesis
is stated and this is then experimentally tested.
a Define ‘hypothesis’.
b How are theories and principles different from a
hypothesis?
2
Which of the following describes an inquiry question?
A If an object is subject to a constant net force, then it
will move with a constant acceleration.
B What features suggest that sound is a mechanical
wave?
C Increasing the voltage in an electric circuit causes
an increase in the current.
D The momentum in an inelastic collision was
conserved.
3
In a practical investigation, a student changes the
voltage by adding or subtracting batteries in series to
the circuit.
a How could the voltage be a discrete variable?
b How could it be a continuous variable?
4
In another experiment a student uses the following
range of values to describe the brightness of a light:
dazzling, bright, glowing, dim, off
What type of variable is ‘brightness’?
5
Select the best hypothesis from the three options
below. Give reasons for your choice.
A Hypothesis 1: If both the angular momentum and
inertia of a rotating system are increased, then the
angular (rotational) velocity will also increase.
B Hypothesis 2: Your position during angular airborne
motion affects your inertia.
C Hypothesis 3: If rotational velocity increases as
radius decreases, then a springboard diver’s
angular (rotational) velocity is slower when they
hold a stretched (layout) position than when they
are in a tuck position, if they take off with the same
angular momentum.
CHAPTER 1 | WORKING SCIENTIFICALLY
9
1.2 Planning investigations
Once you have formulated your hypothesis, you will need to plan and design your
investigation. Taking the time to carefully plan and design a practical investigation
before beginning will help you to maintain a clear and concise focus throughout.
Preparation is essential. This section is a guide to some of the key steps that should
be taken when planning and designing a practical investigation.
WRITING THE METHODOLOGY
The methodology of your investigation is a step-by-step procedure. When detailing
the methodology, ensure it meets the criteria for a valid, reliable and accurate
investigation.
Methodology elements
Validity
Validity means that an experiment or investigation is actually testing the hypothesis
and following the purpose. Will the investigation provide data that is relevant to the
question?
To ensure an investigation is valid, it should be designed so that only one variable
is being changed at a time. The remaining variables must remain constant so that
meaningful conclusions can be drawn about the effect of each variable in turn.
To ensure validity, carefully determine:
• the independent variable; that is, the variable that will be changed and how it will
be changed
• the dependent variable; that is, the variable that will be measured
• the controlled variables; that is, the variables that must remain constant, and how
they will be maintained.
A valid experiment must also gather accurate and reliable results. In other words,
if the data generated is not reliable then the method is not a valid choice for testing
the hypothesis.
FIGURE 1.2.1 Replication increases the
reliability of your investigation. By repeating the
investigation you can average the results and
minimise random errors.
10
Reliability
Reliability means that if an experiment is repeated many times, the results will be
consistent. Reliability can be ensured by:
• defining the control
• ensuring there is sufficient replication of the experiment to minimise error.
It is important to understand the difference between the controlled variables and
the control. Controlled variables are variables that are kept constant in the experiment
so that they do not affect the results. The control is an identical experiment except
that the independent variable is not changed.
A control can be:
• negative: the effect or change is expected in the experimental group but not in
the control
• positive: the effect or change is expected in the control but not in the experimental
group.
The expectations are based on previous experiments or observations. When
the controls do not behave as expected, the data obtained from an experiment or
observation is not reliable.
It is also important to determine how many times the experiment needs to be
replicated (Figure 1.2.1). Many scientific investigations lack sufficient repetition to
ensure that the results can be considered reliable and repeatable. For example, if
you were to only take one reading in an investigation there is a possibility that the
result is reliable, but you would have no way of knowing. You need to repeat the
experiment enough times to ensure that the result is reliable.
CHAPTER 1 | WORKING SCIENTIFICALLY
•
•
•
Repeat readings: Repeat each reading three times, record each measurement
and then average the three measurements. This allows random errors to be
identified. If one reading differs too much from the others, you might have to
discard it before averaging. (This type of reading is called an outlier.) Averaging
your results minimises random error. The different types of error are discussed
in greater detail in Section 1.3.
Sample size: If your experiment involves finding something out about all the
objects in a group, such as the average mass of eggs produced by a farm, you
may not be able to test all the objects. Instead you can test a smaller number of
objects (called a sample) that represent all the objects. If you do this, you need
to make sure your sample contains enough objects to ensure they truly represent
the whole group. The larger the number of objects in your sample, the more
reliable your data will be.
Repeats: If possible repeat the experiment on a different day. Don’t change
anything. If the results are not the same, think about what could have happened.
For example, was the equipment faulty, or were all the controlled variables
correctly identified and kept the same? Repeat the experiment a third time to
confirm which run was correct. More repeats are better; three is a good number
but, if time and resources allow, aim for at least five.
Accuracy and precision
Accuracy refers to the ability to obtain the correct measurement. Precision is the
ability to consistently obtain the same measurement. To obtain precise results, you
must minimise random errors.
Are the instruments to be used sensitive enough? What units will be used? Build
some testing into your investigation to confirm the accuracy and reliability of the
equipment and your ability to read the information obtained.
To understand more clearly the difference between accuracy and precision,
think about firing arrows at an archery target (Figure 1.2.2). Accuracy is being able
to hit the bullseye, whereas precision is being able to hit the same spot every time
you shoot. If you hit the bullseye every time you shoot, you are both accurate and
precise (Figure 1.2.2a). If you hit the same area of the target every time but not the
bullseye, you are precise but not accurate (Figure 1.2.2b). If you hit the area around
the bullseye each time but don’t always hit the bullseye, you are accurate but not
precise (Figure 1.2.2c). If you hit a different part of the target every time you shoot,
you are neither accurate nor precise (Figure 1.2.2d).
(a)
(b)
(c)
GO TO ➤ Section 1.3, page 15
SKILLBUILDER
N
L
Scientific data
All scientists strive to measure
and report accurate and precise
results. However very precise
measurements can be unwieldy —
imagine entering a calculation
with five numbers that were all
measured to 20 decimal places!
Scientists therefore restrict some
measurements to a certain
amount of significant figures or
decimal places.
For example, the speed of
light has been calculated
to be 2.998 × 108 m s−1 to
four significant figures. It is
also commonly written as
3.0 × 108 m s−1, which has been
rounded to two significant figures.
Neither measurement is incorrect,
but 2.998 × 108 m s−1 is the more
precise measurement.
It is important that you are aware
that the reliability of scientific data
can vary, depending on the source.
Always check that the data you
are using has come from a reliable
source.
(d)
FIGURE 1.2.2 Examples of accuracy and precision: (a) both accurate and precise, (b) precise but not
accurate, (c) accurate but not precise, and (d) neither accurate nor precise.
Reasonable steps to ensure the accuracy of an investigation include considering:
the unit in which the independent and dependent variables will be measured
the instruments that will be used to measure the independent and dependent
variables.
Select and use appropriate equipment, materials and methods. For example,
select equipment that can measure in smaller units to reduce uncertainty, and repeat
the measurements to confirm them.
Describe the materials and method in appropriate detail. This should ensure that
every measurement can be repeated and the same result obtained within reasonable
margins of experimental error (less than 5% is reasonable).
•
•
CHAPTER 1 | WORKING SCIENTIFICALLY
11
GO TO ➤ Section 1.3, page 16
Sourcing appropriate materials and technology
When designing your investigation, you will need to decide on the materials,
technology and instrumentation that will be used to carry out your research. It is
important to find the right balance between items that are easily accessible and those
that will give you accurate results. As you move onto conducting your investigation,
it will be important to take note of the precision of your chosen instrumentation and
how this affects the accuracy and validity of your results. This is discussed in more
detail in Section 1.3.
GO TO ➤ Section 1.4, page 19
Data analysis
Data analysis is part of the method. Consider how the data will be presented and
analysed. Preparing an empty table showing the data that needs to be obtained will
help you to plan the investigation.
A wide range of analysis tools are available. For example, tables organise
data so that patterns can be seen, and graphs can show relationships and make
comparisons. The nature of the data being collected, such as whether the variables
are qualitative or quantitative, influences the type of method or tool that you can
use to analyse the data. The purpose and the hypothesis will also influence the
choice of analysis tool.
Data analysis is covered in more detail in Section 1.4.
Modifying the procedure
The procedure (also known as the methodology) may need modifying as the
investigation is carried out. The following actions will help to determine any issues
in the procedure and how to modify them:
• Record everything.
• Be prepared to make changes to the approach.
• Note any difficulties encountered and the ways they were overcome. What
were the failures and successes? Every test carried out can contribute to the
understanding of the investigation as a whole, no matter how much of a disaster
it may seem at first.
• Do not panic. Go over the theory again, and talk to the teacher and other
students. A different perspective can lead to a solution.
If the expected data is not obtained, don’t worry. As long as it can be critically
and objectively evaluated, and the limitations of the investigation are identified and
further investigations proposed, the work is worthwhile.
ETHICAL AND SAFETY GUIDELINES
Ethical considerations
When you are planning an investigation, identify all possible ethical considerations
and consider how to reduce or eliminate them. Ethical issues could include:
• How could this affect wider society?
• Does it involve humans or animals?
• Does one group benefit over another; for example, one individual, a group of
individuals or a community? Is it fair?
• Who will have access to the data and results?
• Does it prevent anyone from gaining their basic needs?
• How can this impact on future ethical decisions or issues? For example, even if
an application is ethical, could it lead to applications that are unethical?
Investigations that involve humans or animals usually require ethics approval.
This includes experiments directly involving humans or animals, as well as public
surveys and other investigations that collect information about people. Ask your
teacher for further information about this issue.
12
CHAPTER 1 | WORKING SCIENTIFICALLY
Risk assessments
It is important for the safety of yourself and the safety of others that the potential
risks are considered when you are planning an investigation.
Everything we do has some risk involved. Risk assessments are performed to
identify, assess and control hazards. A risk assessment should be performed for
any situation, whether in the laboratory or outside in the field. Always identify the
risks and control them to keep everyone safe. For example, carry out electrical
experiments with low DC voltages (e.g. less than 12 volts) coupled to resistors so
that the currents in the circuits are of the order of milliamps. At all times avoid direct
exposure to 240 V AC household voltages (Figure 1.2.3).
To identify risks, think about:
• the activity that will be carried out
• the equipment or materials that will be used.
The following list of risk controls is organised from most effective to least
effective:
1 Elimination: Eliminate dangerous equipment, procedures or substances.
2 Substitution: Find different equipment, procedures or substances to use that will
achieve the same result, but have less risk associated.
3 Isolation: Ensure there is a barrier between the person and the hazard. Examples
include physical barriers such as guards in machines, or fume hoods to work
with volatile substances.
4 Engineering controls: Modify equipment to reduce risks.
5 Administrative controls: Provide guidelines, special procedures, and warning
signs for any participants, and ensure that behaviour is safe.
6 Protective equipment: Wear safety glasses, lab coats, gloves and respirators etc.
where appropriate, and provide these to other participants.
FIGURE 1.2.3 When planning an investigation
you need to identify, assess and control hazards.
Science outdoors
Sometimes investigations and experiments will be carried out outdoors. Working
outdoors has its own set of potential risks and it is equally important to consider
ways of eliminating or reducing these risks.
As an example, read Table 1.2.1, which contains examples of risks associated
with outdoor research.
TABLE 1.2.1
Examples of risks associated with outdoor research.
Risks
Control measures
sunburn
wear sunscreen, a hat and sunglasses; use shade where
possible
hot weather
rest and drink fluids regularly
cold, wind, rain
wear warm, windproof and waterproof clothing
bites and stings
use insect repellent and look out for snakes, wasps and
other dangerous animals
trip hazards
be aware of tree roots, rocks etc.
public safety
create barriers so that people know not to enter the area
First aid
Minimising the risk of injury reduces the chance of requiring first aid assistance.
However, it is still important to have someone with first aid training present during
practical investigations. Always tell the teacher or laboratory technician if an injury
or accident happens.
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13
Personal protective equipment
Everyone who works in a laboratory wears items that help keep them safe. This is
called personal protective equipment (PPE) and includes:
• safety glasses
• shoes with covered tops
• disposable gloves for handling chemicals
• a disposable apron or a lab coat if there is risk of damage to clothing
• ear protection if there is risk to hearing.
1.2 Review
SUMMARY
• The methodology of your investigation is a
step-by-step procedure. When detailing the
methodology, ensure it meets the requirements
for a valid, reliable and accurate investigation.
when you carry out your methodology, you and
others are kept safe. If you have elements of your
investigation which are not safe you will need to
reevaluate your design.
• It is important to determine how many times
the experiment needs to be replicated. Scientific
investigations sometimes lack sufficient
repetitions to ensure that the results are reliable
and repeatable.
• It is important to choose appropriate equipment
for your experiment. This means not only personal
protective equipment (PPE) that will help keep you
safe, but also instrumentation that will give you
accurate results.
• Risk assessments must be carried out before
conducting an investigation to make sure that,
KEY QUESTIONS
1
A journal article reported the materials and method
used in order to conduct an experiment. The
experiment was repeated three times, and all values
were reported in the results section of the article.
Which one of the following is supported by repeating
an experiment and reporting results?
A validity
B reliability
C credibility
D systematic errors
2
3
14
A student wanted to find out whether you can hit a
ball harder with a two-handed grip of the bat instead
of a one-handed grip. What would be the independent
variable for their experiment?
a Explain what is meant by the term ‘controlled
experiment’.
b Using an example, distinguish between independent
and dependent variables.
CHAPTER 1 | WORKING SCIENTIFICALLY
4
You are conducting an experiment to find the time
taken for a swimmer to complete a lap of a pool.
Discuss the accuracy of your results if you are:
a using a stop watch
b watching a clock
c recording the motion with a camera.
5
You are conducting a practical investigation to find the
acceleration due to gravity by dropping a ball from
different heights and measuring the time it takes to fall
to the ground. What sort of risks may be involved in
this investigation?
6
Give the correct term (valid, reliable, or accurate) that
describes an experiment with the following conditions.
a The experiment addresses the hypothesis and
purposes.
b The experiment is repeated and consistent results
are obtained.
c Appropriate equipment is chosen for the desired
measurements.
1.3 Conducting investigations
Once the planning and design of a practical investigation is complete, the next
step is to undertake the investigation and record the results. As with the planning
stages, there are key steps and skills to keep in mind to maintain high standards and
minimise potential errors throughout the investigation (Figure 1.3.1).
This section will focus on the best methods of conducting a practical investigation,
by systematically generating, recording and processing data.
COLLECTING AND RECORDING DATA
For an investigation to be scientific, it must be objective and systematic. Ensuring
familiarity with the methodology and protocols before beginning will help you to
achieve this.
When working, keep asking questions. Is the work biased in any way? If changes
are made, how will they affect the study? Will the investigation still be valid for the
purpose and hypothesis?
It is essential that during the investigation the following are recorded in the
logbook:
• all quantitative and qualitative data collected
• the methods used to collect the data
• any incident, feature or unexpected event that may have affected the quality or
validity of the data.
The data recorded in the logbook is the raw data. Usually this data needs to
be processed in some manner before it can be presented. If an error occurs in the
processing of the data or you decide to present the data in an alternative format, the
recorded raw data will always be available for you to refer back to.
Safe work practices
Remember to always employ safe work practices while conducting your experiment.
See Section 1.2 for how to conduct risk assessments.
You will also need to keep in mind safe procedures to follow when disposing of
waste. This will depend on the types of waste produced throughout your experiment.
Your teacher will be able to direct you on how best to approach waste disposal.
Education or government websites can also be a great source of information.
FIGURE 1.3.1 When carrying out your
investigation try to maintain high standards
to minimise potential errors.
GO TO ➤ Section 1.2, page 13
IDENTIFYING ERRORS
Most practical investigations have errors associated with them. Errors can occur for a
variety of reasons. Being aware of potential errors helps you to avoid or minimise them.
For an investigation to be accurate, it is important to identify and record any errors.
There are three types of errors:
• mistakes (avoidable errors)
• systematic errors
• random errors.
Types of error
Mistakes
Mistakes are avoidable errors. For example, mistakes made during water quality
analysis could include:
• misreading the numbers on a scale
• not labelling a sample adequately
• spilling a portion of a sample.
A measurement that involves a mistake must be rejected and not included in any
calculations, or averaged with other measurements of the same quantity. Mistakes
are often not referred to as errors because they are caused by the experimenter
rather than the experiment or the experimental method.
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15
Systematic errors
Systematic errors are errors that are consistent and will occur again if the
investigation is repeated in the same way. They are usually a result of instruments
that are not calibrated correctly or methods that are flawed.
An example of a systematic error would be if a ruler mark indicating 5 cm was
actually at 4.9 cm because of a manufacturing error. Another example would be
if the researcher repeatedly used a piece of equipment that was not calibrated
correctly throughout the investigation. Figure 1.3.2 shows how traffic police reduce
systematic errors in their data collection.
FIGURE 1.3.2 To avoid a systematic error, make
sure that you are using measuring equipment
correctly. Laser speed guns, for example, are
held so that the aim is kept on a single target
point for the duration of the reading.
Random errors
Random errors occur in an unpredictable manner and are generally small. A
random error could be, for example, the result of a researcher reading the same result
correctly one time and incorrectly another time, or an instrument not functioning
correctly because of a power failure or low battery power. If a controlled variable is
not kept constant throughout the investigation, this can also be the cause of random
errors.
Techniques to reduce error
Designing the method carefully, including selection and use of equipment, will help
reduce errors.
Appropriate equipment
Use the equipment that is best suited to the data that needs to be collected to validate
the hypothesis. Determining the units of the data being collected and at what scale
will help to select the correct equipment. Using the right unit and scale will ensure
that measurements are more accurate and precise (with smaller systematic errors).
Significant figures are the numbers that convey meaning and precision. The
number of significant figures used depends on the scale of the instrument. It is
important to record data to the number of significant figures available from the
equipment or observation. Using either a greater or smaller number of significant
figures can be misleading.
Review the following examples to learn more about significant figures:
• 15 has two significant figures
• 3.5 has two significant figures
• 3.50 has three significant figures
• 0.037 has two significant figures
• 1401 has four significant figures.
Although digital scales can measure to many more than two figures and
calculators can show up to 12 figures, be sensible and follow the significant figure
rules. Do not round off your answer until you no longer need it for subsequent
calculations. Here is an example:
To calculate gravitational potential energy (U), the formula is ΔU = mgΔh.
If g = 9.81 m s−2, m = 7.50 kg, and h = 0.64 m (64 cm) then:
U = 9.81 × 7.50 × 0.64 = 47.09 J
However, you can only give the answer to the least number of significant figures
in the data, which in this case is two for h, so U = 47 J.
Calibrated equipment
Some equipment, such as some motion sensors, needs to be calibrated before use
to account for the temperature at the time. Before carrying out the investigation,
make sure the instruments or measuring devices are properly calibrated and are
functioning correctly. For example, measure the temperature and apply a correction
to the speed of sound to calibrate a motion sensor if necessary. Making sure your
equipment is correctly calibrated will help to ensure accuracy in your data.
16
CHAPTER 1 | WORKING SCIENTIFICALLY
Correct use of equipment
Use the equipment properly. Ensure any necessary training has been done to use the
equipment and that you have had an opportunity to practise using the equipment
before beginning the investigation. Improper use of equipment can result in
inaccurate, imprecise data with large errors, and the validity of the data can be
compromised (Figure 1.3.3).
Incorrect reading of measurements is a common misuse of equipment. Make
sure all of the equipment needed in the investigation can be used correctly, and
record the instructions in detail so they can be referred to if the data doesn’t appear
to be correct.
Repeat the investigation
As discussed in Section 1.2, repeating the investigation and averaging the results will
generate data that is more reliable. Modifications to your procedure may need to
be looked at before repeating the investigation to ensure that all variables are being
tested under the same conditions.
Using information from secondary sources
As you conduct your investigation, it is important to note any information you use
that has come from secondary sources. This must then be stated in your written
report. This is discussed in more detail in Section 1.7.
Categorising the information and evidence you find while you are researching
will make it easier to locate information later and to write your final investigation.
Categories you might use while researching could include:
• research methods
• key findings
• evidence
• research relevance
• use
• future concerns.
Record information from resources in a clear way so you can retrieve it and use
it. The style used to record the resources you use is described in Section 1.7.
GO TO ➤ Section 1.2, page 10
GO TO ➤ Section 1.7, page 37
GO TO ➤ Section 1.7, page 39
FIGURE 1.3.3 In order to produce accurate data, make sure all your equipment is calibrated correctly
and you have taken the time to understand how to use it safely and correctly.
CHAPTER 1 | WORKING SCIENTIFICALLY
17
1.3 Review
SUMMARY
• Random errors occur in an unpredictable manner
and are generally small. A random error could be,
for example, the result of a researcher reading
the same result correctly one time and incorrectly
another time.
• It is essential that during the investigation, the
following are recorded:
-- all quantitative and qualitative data collected
-- the methods used to collect the data
-- any incident, feature or unexpected event that
may have affected the quality or validity of
the data.
• The number of significant figures used depends
on the accuracy of the measurements. It is
important to record data to the number of
significant figures available from the equipment
or observation.
• A systematic error is an error that is consistent
and will occur again if the investigation is repeated
in the same way. Systematic errors are usually
a result of instruments that are not calibrated
correctly or methods that are flawed.
KEY QUESTIONS
1
Both sets of data below contain errors. Identify which
set is more likely to contain a systematic error and
which is more likely to contain a random error.
5
A scientist carries out a set of experiments, analyses
the results and publishes them in a scientific journal.
Other scientists in different laboratories repeat the
experiment, but do not get the same results as the
original scientist. Suggest several reasons that could
explain this.
6
The masses of 1 cm3 cubes of potato were recorded,
then the cubes were placed in distilled water. After
60 minutes, the cubes were weighed again and the
difference in mass was calculated. What type of error
is involved:
a if the electronic scales only measured in
1 g increments?
b if the electronic scales were affected briefly by
a power surge?
Data set A: 11.4, 10.9, 11.8, 10.6, 1.5, 11.1
Data set B: 25, 27, 22, 26, 23, 25, 27
18
2
What type of error is associated with:
a inaccurate measurements?
b imprecise measurements?
3
It is possible to calculate the speed of a moving object
by dividing the distance travelled by the time taken.
You are able to run 250 m in 16.67 s; calculate your
speed to the correct number of significant figures.
4
If you use a value of 7.50 kg for mass and 1.4 m s−1 for
speed in a calculation, what would be the appropriate
number of significant figures in the answer?
CHAPTER 1 | WORKING SCIENTIFICALLY
1.4 Processing data and information
Once you have conducted your investigation and collected data, you will need
to find the best way of presenting and analysing it. This section is a guide to the
different forms of representation that will help you to better understand your data.
RECORDING AND ORGANISING QUANTITATIVE DATA
Raw data is unlikely to be used directly to evaluate the hypothesis. However, raw
data is essential to the investigation and plans for collecting the raw data should
be made carefully. Consider the formulas or graphs that will be used to analyse
the data at the end of the investigation. This will help to determine the type of raw
data that needs to be collected (and the equipment needed) in order to evaluate the
hypothesis.
For example, to calculate take-off velocity for a vertical jump, three sets of raw
data will need to be collected: the athlete’s standing body weight (using scales), the
ground reaction force (using a force platform) and the time between take-off and
landing (using a stopwatch). The data can then be processed to obtain the take-off
velocity.
Once you have determined the data that needs to be collected, prepare a table in
which to record the data.
ANALYSING AND PRESENTING DATA
The raw data that has been obtained needs to be presented in a way that is clear,
concise and accurate.
There are a number of ways of presenting data, including tables, graphs, flow
charts and diagrams. The best way of visualising the data depends on its nature.
Try several formats before making a final decision, to create the best possible
presentation.
Presenting raw and processed data in tables
Tables organise data into rows and columns and can vary in complexity according
to the nature of the data. Tables can be used to organise raw data and processed data
or to summarise results.
The simplest form of a table has two columns. The left column contains the
independent variable (the one being changed) and the right column contains
the dependent variable (the one that may change in response to a change in the
independent variable).
Tables should have the following features:
• a descriptive title
• column headings (including the unit)
• aligned numbers (align the decimal points)
• the independent variable in the left column
• the dependent variable in the right column.
Look at the table in Figure 1.4.1, which has been used to organise raw and
processed data about the effect of current on voltage.
CHAPTER 1 | WORKING SCIENTIFICALLY
19
Effect of current on voltage
clear title
Sample
Current (A)
Voltage (V)
1
2
3
4
5
6
7
8
0.05
0.05
0.04
0.04
0.03
0.03
0.02
0.02
1.81
1.56
1.42
1.24
1.05
0.93
0.76
0.63
Resistance
(Ω or V A–1)
independent
variable
dependent
variable
replicates
grouped
together
36.20
31.20
35.50
31.00
35.00
31.00
38.00
31.50
heading for each column
(units in brackets)
consistent use of
significant figures
FIGURE 1.4.1 A simple table listing the raw data obtained in the second and third columns and
processed data in the fourth column
A table of processed data usually presents the average values of the data, called
the mean. However, the mean on its own does not provide an accurate picture of
the results.
To report processed data more accurately, the uncertainty should be presented
as well.
Uncertainty
When there is a range of measurements of a particular value, the mean must be
accompanied by the uncertainty for your results to be presented as a mean in an
accurate way. In other words, the mean must be accompanied by a description of
the range of data obtained.
Uncertainty is calculated by:
uncertainty = ±(maximum difference from the mean)
Worked example 1.4.1
CALCULATING UNCERTAINTY
The speeds, in km h−1, of cars travelling down a certain road were:
46, 50, 55, 48, 50, 58, 45
Find the mean speed and the uncertainty for these values.
Thinking
Working
Calculate the mean speed.
Mean =
(46 + 50 + 55 + 48 + 50 + 58 + 45) ÷ 7
= 50.3 = 50 km h−1
Calculate the maximum
difference from the mean.
Maximum difference is 58 − 50 = 8, so the
uncertainty is 8.
Write out the mean speed
and include the uncertainty.
Mean speed is 50 ± 8 km h−1.
Worked example: Try yourself 1.4.1
CALCULATING UNCERTAINTY
James is practising his tennis serve. The speed of the ball, in m s−1, is measured to be:
45, 52, 51, 49, 49, 53, 47
Find the mean and uncertainty for these values.
20
CHAPTER 1 | WORKING SCIENTIFICALLY
Absolute uncertainty
The absolute uncertainty is related to the precision of the instrument being used.
Absolute uncertainty is equal to half the smallest unit of measurement. For
example, if you are measuring the length of an object with a ruler on which the
smallest units are millimetre markings, you will be able to measure to the nearest
millimetre. The true value of the measurement could then be anywhere within half
a millimetre of your observed length, so the absolute uncertainty in this case would
be ±0.5 mm.
Percentage uncertainty
Percentage uncertainty, also known as relative uncertainty, is another way of
stating how precise a measurement is. To calculate the percentage uncertainty
take the absolute uncertainty and divide it by the measurement, then multiply by
100. For example, if you have measured a distance of 230 cm using a ruler and the
absolute uncertainty for this measurement is ±0.5 cm, then:
0.5 × 100 = ± 0.2%
Percentage uncertainty = 230
Absolute uncertainty = ± 12 × precision
uncertainty
Percentage uncertainty = absolute
measured value × 100
Absolute and percentage uncertainty are sometimes referred to as absolute and
percentage error.
Other descriptive statistics measures
The mean and the uncertainty are statistical measures that help describe data
accurately. Other statistical measures that can be used, depending on the data
obtained, are:
• mode: the mode is the value that appears most often in a data set. This measure
is useful to describe qualitative or discrete data. For example, the mode of the set
of values [0.01, 0.01, 0.02, 0.02, 0.02, 0.03, 0.04] is 0.02.
• median: the median is the ‘middle’ value of an ordered list of values. It is the
value that has equal numbers of values to its left and to its right. For example, the
median of the set of values [5, 5, 8, 8, 9, 10, 20] is 8. The median is used when
the data range is spread, which may happen if unusually large or small results
make the mean unreliable.
Graphs
In general, tables provide more detailed data than graphs, but it is easier to observe
trends and patterns in data in graphical form than in tabular form.
Graphs are used when two variables are being considered and one variable is
dependent on the other. The graph shows the relationship between the variables.
There are several types of graphs that can be used, including line graphs, bar
graphs and pie charts. The best one to use will depend on the nature of the data.
General rules to follow when making a graph include the following:
• Keep the graph simple and uncluttered.
• Use a descriptive title.
• Represent the independent variable on the x-axis and the dependent variable on
the y-axis.
• Make axes proportionate to the data. For example, your y-axis data might range
from 0 to 100 and your x-axis data might be between 1 and 5. So you could
divide the y-axis into 10 units of 10 and your x-axis into 5 units of 1.
• Clearly label axes with both the variable and the unit in which it is measured.
CHAPTER 1 | WORKING SCIENTIFICALLY
21
Graph of velocity of glider with time as
it travels down an inclined air track
Always give a descriptive title
to the graph or diagram.
40
Always mark where axis value is.
Velocity (cm s–1)
30
Draw line of best fit.
Use a ruler for straight
lines and a practised
sweep for curves.
Always write
values clearly.
20
If you have an outlier go
back to your workings
and check it again to see
if you can explain why.
10
Origin does not have to be (0, 0).
0
0
0.5
Different scales
may be used.
FIGURE 1.4.2
1.0
1.5
2.0
2.5
3.0
Time (s)
Give the quantity name
in full with units in ( ).
A graph shows the relationship between two variables.
Line graphs
Line graphs (Figure 1.4.2) are a good way of representing continuous quantitative
data. In a line graph, the values are plotted as a series of points on the graph. There
are two ways of joining these points:
• A line can be ruled from each point to the next, as shown in Figure 1.4.3(a).
This only shows the overall trend; it does not predict the value of the points
between the plotted data. This type of graph is used when there is no obvious
trend between the variables.
• The points can be joined with a single smooth straight or curved line, as shown
in Figure 1.4.3(b). This creates a trend line, also known as a line of best fit. The
trend line does not have to pass through every point but should go close to as
many points as possible. It is used when there is an obvious trend between the
variables. A trend line can be linear or curved.
22
CHAPTER 1 | WORKING SCIENTIFICALLY
Height of river (m)
3
Height of river at bridge flood gauge
2
1
0
J F M A M J J A S O N D
Months 2010
(b)
Stopping distance (m)
(a)
Stopping distance for car
50
40
30
20
10
0
0
20
40
60
Speed of car (km h−1)
80
FIGURE 1.4.3 (a) The data in the graph is joined from point to point. (b) The data in the graph is
joined with a line of best fit, which shows the general trend.
Outliers
Sometimes there may be one data point that does not fit with the trend and is clearly
an error. This is called an outlier. An outlier is often caused by a mistake made in
measuring or recording data, or from a random error in the measuring equipment. If
there is an outlier, include it on the graph and label it as an outlier, but ignore it when
adding a trend line. This is shown in Figure 1.4.2, where the point (1.5, 6) is an outlier.
1.4 Review
SUMMARY
• Consider how the data will be presented and
analysed. A wide range of analysis tools could be
used. For example, tables organise data so that
patterns can be established and graphs can show
relationships and comparisons.
by the uncertainty, for your results to be presented
as a mean in an accurate way.
• General rules to follow when making a graph
include the following:
-- Keep the graph simple and uncluttered.
• The simplest form of a table is a two-column
format in which the first column contains the
independent variable (the one being changed)
and the second column contains the dependent
variable (the one that may change in response to
a change in the independent variable).
-- Use a descriptive title.
-- Represent the independent variable on the
x-axis and the dependent variable on the y-axis.
-- Make axes proportionate to the data.
-- Clearly label axes with both the variable and the
unit in which it is measured.
• When there is a range of measurements of a
particular value, the mean must be accompanied
KEY QUESTIONS
1
Define the term ‘outlier’.
2
For the set of numbers 21, 28, 19, 19, 25, 24,
determine:
a the mean
b the mode
c the median.
3
Plot the following data set, assigning each variable to
the appropriate axis on the graph.
Current (A)
0.06
0.05
0.04
0.03
0.02
Voltage (V)
2.07
1.56
1.24
0.93
0.63
4
How can the general pattern (trend) of a graph be
represented once the points are plotted?
5
State the mean and uncertainty for the following data:
33, 36, 28, 37, 29, 27, 30, 31
6
In conducting an experiment comparing the speed
of sound to air temperature, your thermometer has
units of 1°C and you have found the air temperature
to be 20°C. Calculate:
a the absolute uncertainty
b the percentage uncertainty.
CHAPTER 1 | WORKING SCIENTIFICALLY
23
1.5 Analysing data and information
Now that the chosen topic has been thoroughly researched, the investigation has
been conducted and data collected, it is time to draw it all together. You will now
need to analyse your results to better understand the physical processes behind
them.
FIGURE 1.5.1
To discuss and conclude your investigation, utilise the raw and processed data.
EXPLAINING RESULTS IN THE DISCUSSION
The discussion is the part of the investigation where the evaluation and explanation
of the investigation methods and results takes place. It is the interpretation of what
the results mean.
The key sections of the discussion are:
• analysing and evaluating data
• evaluating the investigative method
• explaining the link between the investigation findings and the relevant physics
concepts.
Consider the message to be conveyed to the audience, when writing the discussion.
Statements need to be clear and concise. At the conclusion of the discussion, the
audience must have a clear idea of the context, results and implications of the
investigation.
ANALYSING AND EVALUATING DATA
In the discussion, the findings of the investigation need to be analysed and
interpreted.
• State whether a pattern, trend or relationship was observed between the
independent and dependent variables. Describe what kind of pattern it was and
specify under what conditions it was observed.
• Were there discrepancies, deviations or anomalies in the data? If so, these should
be acknowledged and explained.
• Identify any limitations in the data you have collected. Perhaps a larger sample
or further variations in the independent variable would lead to a stronger
conclusion.
24
CHAPTER 1 | WORKING SCIENTIFICALLY
Trends in line graphs
Graphs are drawn to show the relationship, or trend, between two variables, as
shown in Figure 1.5.2.
• Variables that change in linear or direct proportion to each other produce a
straight, sloping trend line.
• Variables that change exponentially in proportion to each other produce a
curved trend line.
• When there is an inverse relationship, one variable increases as the other variable
decreases.
• When there is no relationship between two variables, one variable will not change
even if the other changes.
y
y
x
x
Direct or linear proportional
relationship
Inverse direct or linear
proportional relationship
• Variables change at the same
rate (graph line is straight,
slope is constant).
• Positive relationship—as
x increases, y increases.
• Variables change at the same
rate (graph line is straight,
slope is constant).
• Negative relationship—as
x increases, y decreases.
y
y
x
x
Exponential relationship
Inverse exponential relationship
• As x increases, y increases
slowly, then more rapidly.
• As x increases, y decreases
rapidly, then more slowly, until
a minimum y value is reached.
y
y
x
Exponential rise, then levels off
or plateaus (stops rising)
• As x increases, y increases
rapidly at first, then slows,
then finally does not increase
at all—y reaches a maximum value.
FIGURE 1.5.2
x
No relationship between x and y
• As x increases, y remains the same.
Various relationships can exist between two variables.
CHAPTER 1 | WORKING SCIENTIFICALLY
25
Remember that the results may be unexpected. This does not make the
investigation a failure. However, the findings must be related to the hypothesis,
purpose and method.
Mathematical models
SKILLBUILDER
N
Graphing a linear
relationship
When analysing data from a linear
relationship, it is first necessary to
obtain a graph of the data and an
equation for the line that best fits
the data.
If you are plotting your graph
manually on paper then proceed
as follows:
1 Plot each data point on clearly
labelled, unbroken axes.
2 Identify and label but otherwise
ignore any suspect data points.
3 Draw, by eye, a ‘line of best
fit’ for the points. The points
should be evenly scattered
either side of the line.
After analysing your data using tables and graphs, it might be possible to find a
mathematical relationship to describe your results. For example, your graph may
produce a straight line, so there is some sort of linear relationship between the two
variables.
Linear relationships
Some relationships studied in physics are linear (that is, a straight line) while others
are not. It is possible to manipulate non-linear data so that a linear graph reveals a
measurement. Linear relationships and their graphs are fully specified with just two
numbers: gradient, m, and vertical axis intercept, c. In general, linear relationships
are written:
y = mx + c
The gradient m can be calculated from the coordinates of two points on the line:
rise
m = run
=
y2 − y1
x2 − x1
where (x1, y1) and (x2, y2) are any two points on the line. Don’t forget that m
and c have units. Omitting these is a common error.
Worked example 1.5.1
FINDING A LINEAR RELATIONSHIP FROM DATA
4 Locate the vertical axis intercept
and record its value as c.
5 Choose two points on the
line of best fit to calculate the
gradient. Do not use two of the
original data points as this will
not give you the gradient of the
line of best fit.
6 Write y = mx + c, replacing
x and y with appropriate
symbols, and use this equation
for any further analysis.
Some students used a computer with an ultrasonic detector to obtain
the speed–time data for a falling tennis ball. They wished to measure the
acceleration of the ball as it fell. They assumed that the acceleration was nearly
constant and that the relevant relationship was v = u + at, where v is the speed of
the ball at any given time, u was the speed when the measurements began, a is
the acceleration of the ball and t is the time since the measurement began.
Their computer returned the following data:
Time (s)
Speed (m s−1)
0.0
1.25
0.1
2.30
0.2
3.15
0.3
4.10
0.4
5.25
0.5
6.10
0.6
6.95
Find their experimental value for acceleration.
26
CHAPTER 1 | WORKING SCIENTIFICALLY
Thinking
Working
Decide which axes each
of your variables should
be on.
The independent variable is the speed, so this will go
on the y-axis.
The dependent variable is the time, so this will go on
the x-axis.
Speed (m s−1)
Graph your data as a
scatter plot and draw a
line of best fit.
Find the equation for the
line of best fit.
y = 9.5714x + 1.2857
8
7
6
5
4
3
2
1
0
0.00
0.20
0.40
Time (s)
0.60
0.80
This graph of the data was created on a computer
spreadsheet. The line of best fit was created
mathematically and plotted. The computer
calculated the equation of the line. Graphics
calculators can also do this.
v = 9.5714t + 1.2857
Write out your linear
relationship in the form
required.
v = 9.5714t + 1.2857
If this is rearranged and the constants are rounded
off, the equation is:
v = 1.3 + 9.6t.
The acceleration is 9.6 m s−2.
State the answer.
Worked example: Try yourself 1.5.1
FINDING A LINEAR RELATIONSHIP FROM DATA
The downward force is measured for a variety of different masses.
Mass (kg)
Force (N)
0.25
2.4
0.50
4.9
0.75
7.4
1.0
9.9
1.3
12.8
Find the linear relationship between the values for mass m and force F.
Manipulating non-linear data
Suppose you were examining the relationship between two quantities B and d and
had good reason to believe that the relationship between them is:
B = dk
where k is some constant value.
CHAPTER 1 | WORKING SCIENTIFICALLY
27
Clearly, this relationship is non-linear and a graph of B against d will not be a
straight line. By thinking about the relationship it can be seen that in ‘linear form’:
B = k d1
↑ ↑↑
y = mx + c
The graph of B (on the vertical axis) against d1 (on the horizontal axis) will be
linear. The gradient of the line will be k and the vertical intercept c will be zero. The
line of best fit would be expected to go through the origin because, in this case, there
is no constant added and so c is zero.
In the above example, the graph of the raw data would just show that B is larger
as d is smaller. It would be impossible to determine the mathematical relationship
just by looking at a graph of the raw data.
A graph of raw data will not give the mathematical relationship between the
variables but can give some clues. The shape of the graph of raw data may suggest a
possible relationship. Several relationships may be tried and then the best is chosen.
Once this is done, it is not proof of the relationship but, possibly, strong evidence.
When an experiment involves a non-linear relationship, the following procedure
is followed:
1 Plot a graph of the original raw data.
2 Choose a possible relationship based on the shape of the initial graph and your
knowledge of various mathematical and graphical forms.
3 Work out how the data must be manipulated to give a linear graph.
4 Create a new data table.
Then follow the steps given in the SkillBuilder on page 26. It may be necessary
to try several mathematical forms to find one that seems to fit the data.
Worked example 1.5.2
FINDING A NON-LINEAR RELATIONSHIP FROM DATA
Some students were investigating the relationship between current and
resistance for a new solid-state electronic device. They obtained the data shown
in the table.
Current, I (A)
Resistance, R (Ω)
1.5
22
1.7
39
2.2
46
2.6
70
3.1
110
3.4
145
3.9
212
4.2
236
According to the theory they had researched on relevant internet sites, the
students believed that the relationship between I and R is:
R = dI3 + g
where d and g are constants.
By appropriate manipulation and graphical techniques, find the students’
experimental values for d and g.
28
CHAPTER 1 | WORKING SCIENTIFICALLY
Thinking
Working
Plot a graph of the
raw data
Resistance (Ω)
250
200
150
100
50
0
0
1
2
3
Current (A)
4
5
It might be argued that the second piece of data is
suspect. The rest of this solution supposes the students
chose to ignore this piece of data.
Work out what you
would have to graph
to get a straight line.
You can see what to graph if you think of the equation
like this:
R = d I3 + g
↑
↑ ↑
↑
y = m x + c
A graph of R on the vertical axis and I 3 on the
horizontal axis would have a gradient equal to d and
a vertical axis intercept equal to g.
Make a new table of
the manipulated data.
The data is manipulated by finding the cube of each of
the values for current.
Current cubed, I3 (A3)
Resistance, R (Ω)
3.38
22
10.65
46
17.58
70
29.79
110
39.30
145
59.32
212
74.09
236
Plot the graph of
manipulated data.
y = 3.1412x + 15.092
Resistance (Ω)
300
250
200
150
100
50
0
0
20
40
Current cubed (A3)
60
80
CHAPTER 1 | WORKING SCIENTIFICALLY
29
Calculate the line of
best fit.
This graph of the data was created on a computer
spreadsheet. The line of best fit was created
mathematically and plotted. The computer calculated
the equation of the line. Graphics calculators can also
do this.
y = 3.1x + 15.1
Find the equation
relating I and R.
The regression line has the equation y = 3.1x + 15.1,
so the equation relating I and R is
R = 3.1I3 + 15.1.
Write out the
values for d and g.
Remember to include
the correct units.
d = 3.1 Ω A−3
g = 15.1 Ω
Worked example: Try yourself 1.5.2
FINDING A NON-LINEAR RELATIONSHIP FROM DATA
Some students were investigating the relationship between distance and the
intensity of sound. They obtained the data shown in the table.
Distance, r (m)
Intensity, I (W m−2)
1
0.04
2
0.01
3
0.005
4
0.003
5
0.002
According to the theory they had researched on relevant internet sites, the
students believed that the relationship between I and r is:
I = P2
r
where P is some constant.
By appropriate manipulation and graphical techniques, find the students’
experimental value for P.
EVALUATING THE METHOD
It is important to discuss the limitations of the investigation methodology. Evaluate
the method and identify any issues that could have affected the validity, accuracy or
reliability of the data. Sources of errors and uncertainty must also be stated in the
discussion.
Once any limitations or problems in the methodology have been identified,
recommend improvements on how the investigation could be conducted if repeated;
for example, suggest how bias could be minimised or eliminated.
Bias
Bias may occur in any part of the investigation method, including sampling and
measurements.
Bias is a form of systematic error resulting from the researcher’s personal
preferences or motivations. There are many types of bias, including:
• poor definitions of both concepts and variables (for example, classifying cricket
pitch surfaces and their interaction with the ball according to resilience without
defining ‘slow’ and ‘fast’)
30
CHAPTER 1 | WORKING SCIENTIFICALLY
•
incorrect assumptions (for example, that footwear type, model and manufacturer
do not affect ground reaction forces, and as a result failing to control these variables
during an investigation on slip risk on different indoor and outdoor surfaces)
• errors in the investigation design and procedure (for example, taking a sample
of a particular group of athletes that samples one particular gender more than
the other in the group).
Some biases cannot be eliminated, but should at least be addressed in the
discussion.
Accuracy
In the discussion, evaluate the degree of accuracy of the measurements for each
variable of the hypothesis. Comment on the uncertainties obtained.
When relevant, compare the chosen method with any other methods that might
have been selected, evaluating the advantages and disadvantages of the selected
method and the effect on the results.
Reliability
When discussing the results, indicate the range of the data obtained from replicates.
Explain how the sample size was selected. Larger samples are usually more reliable,
but time and resources might have been scarce. Discuss whether the results of the
investigation have been limited by the sample size.
The control group is important to the reliability of the investigation. A control
group helps determine if a variable that should have been controlled has been
overlooked and may explain any unexpected results.
Error
Discuss any source of systematic or random error and suggest ways of improving
the investigation.
CRITICALLY EVALUATING SECONDARY SOURCES
Not all sources are credible. It is essential to critically evaluate the content and its
origin. Questions you should always ask when evaluating a source include:
• Who created this message? What are the qualifications, expertise, reputation
and affiliation of the authors?
• Why was it written?
• Where was the information published?
• When was the information published?
• How often is the information referred to by other researchers?
• Are conclusions supported by data or evidence?
• What is implied?
• What is omitted?
• Are any opinions or values being presented in the piece?
• Is the writing objectively and accurately describing a scientific concept or
phenomenon?
• How might other people understand or interpret this message differently from me?
When evaluating the validity or bias of websites, consider its domain extension:
• .gov stands for government
• .edu stands for education
• .org stands for a non-profit organisation
• .com stands for commercial/business.
FIGURE 1.5.3 Honest evaluation and reflection
play important roles in analysing methodology.
CHAPTER 1 | WORKING SCIENTIFICALLY
31
1.5 Review
SUMMARY
• After completing your investigation you will need
to analyse and interpret your data. A discussion of
your results is required.
• It is important to discuss the limitations of the
investigation method. Evaluate the method and
identify any issues that could have affected the
validity, accuracy, precision or reliability of the
data. Sources of errors and uncertainty must also
be stated in the discussion.
-- State whether a pattern, trend or relationship
was observed between the independent and
dependent variables. Describe what kind
of pattern it was and specify under what
conditions it was observed.
• When discussing the results, indicate the range
of the data obtained from replicates. Explain how
the sample size was selected. Larger samples are
usually more reliable, but time and resources are
likely to have been scarce. Discuss whether the
results of the investigation have been limited by
the sample size.
-- If possible, create a mathematical model to
describe your data.
-- Were there discrepancies, deviations or
anomalies in the data? If so, these should be
acknowledged and explained.
-- Identify any limitations in the data collected.
Perhaps a larger sample or further variations
in the independent variable would lead to a
stronger conclusion.
KEY QUESTIONS
32
1
What relationship between the variables is indicated by
a sloping linear graph?
4
What might cause a sample size to be limited in an
investigation?
2
What relationship exists if one variable decreases as
the other increases?
5
3
What relationship exists if both variables increase or
both decrease at the same rate?
After analysing the motion of a falling tennis ball, you
create a mathematical model to describe the speed of
the ball as a function of time: y = 1.3 + 9.6t. Describe
what each of the values in this equation represents.
CHAPTER 1 | WORKING SCIENTIFICALLY
1.6 Problem solving
Having analysed your results you can then apply them to physics concepts in order
to evaluate your conclusions. In this section you will learn how analysing your
investigation leads to a better understanding of the underlying scientific principles
of your research.
DISCUSSING RELEVANT PHYSICS CONCEPTS
To make the investigation more meaningful, it should be explained within the right
context, meaning the related physics ideas, concepts, theories and models. Within
this context, explain the basis for the hypothesis.
For example, if you are studying the impact of temperature on the linear strain
of a material (e.g. a rubber band), some of the contextual information to include in
the discussion could be:
• the definition of linear strain
• the functions of linear strain
• the relationship between linear strain and temperature
• the definitions of material behaviour such as plastic and elastic
• the factors known to affect linear strain
• existing knowledge on the role of temperature on linear strain
• the ranges of temperatures investigated and the reason these temperatures were
chosen
• the materials studied and the reasons for this choice
• methods of measuring the linear strain of a material.
Relating your findings to a physics concept
During the analysing stage of your investigaton (Section 1.5) you were able to find
trends, patterns and mathematical models of your results. This is the framework
needed in which to discuss whether the data supported or refuted the hypothesis.
Ask questions such as:
• Was the hypothesis supported?
• Has the hypothesis been fully answered? (If not, give an explanation of why
this is so and suggest what could be done to either improve or complement the
investigation.)
• Do the results contradict the hypothesis? If so, why? (The explanation must be
plausible and must be based on the results.)
Providing a theoretical context also enables comparison of the results with
existing relevant research and knowledge. After identifying the major findings of
the investigation, ask questions such as:
• How do the findings fit with the literature?
• Do the findings contradict the literature?
• Do the findings fill a gap in the literature?
• Do the findings lead to further questions?
• Can the findings be extended to another situation?
Be sure to discuss the broader implications of the findings. Implications are
the bigger picture. Outlining them for the audience is an important part of the
investigation. Ask questions such as:
• Do the findings contribute to or impact on the existing literature and knowledge
of the topic?
• Are there any practical applications for the findings?
GO TO ➤ Section 1.5, page 24
CHAPTER 1 | WORKING SCIENTIFICALLY
33
DRAWING EVIDENCE-BASED CONCLUSIONS
A conclusion is usually a paragraph that links the collected evidence to the hypothesis
and provides a justified response to the research question.
Indicate whether the hypothesis was supported or refuted and the evidence on
which this is based (that is, the results). Do not provide irrelevant information. Only
refer to the specifics of the hypothesis and the research question and do not make
generalisations.
Read the examples of conclusions for the following hypothesis and research
question.
Hypothesis: If linear deformation (change in length) has a positive relationship
with temperature, then an increase in temperature will cause an increase in linear
deformation.
• Poor response to the hypothesis: Linear deformation has value y1 at temperature
t1 and value y2 at temperature t2.
• Better response to the hypothesis: An increase in temperature from t1 to t2
produces an increase in linear deformation of x in the rubber band.
Inquiry question: Does temperature affect the maximum linear deformation the
material can withstand?
• Poor response to the research question: The results show that temperature does
affect the maximum deformation of a material.
• Better response to the research question: Analysis of the results of the effect of
an increase in temperature from t1 to t2 in the rubber band supports current
knowledge on the effect that an increase in temperature has on increasing
maximum linear deformation.
INTERPRETING SCIENTIFIC AND MEDIA TEXTS
GO TO ➤ Section 1.4, page 31
Sometimes you may be required to investigate claims and conclusions made by
other sources, such as scientific and media texts. As discussed in Section 1.4, some
sources are more credible than others. Once you have analysed the validity of the
secondary source, you will be able to follow the same steps described above in
evaluating their conclusions in order to solve scientific problems.
MODELS
FIGURE 1.6.1
system.
34
A physical model of the solar
Scientific models are used to create and test theories and explain concepts. Different
types of models can be used to study systems, such as the motion of planets within
our solar system (Figure 1.6.1). However, every model has limitations on the type
of information it can provide. For example, the model in Figure 1.6.1 does not show
the relative distances of the planets from the Sun, or the relative sizes of the planets
and the Sun.
Models are created to answer specific questions. How a model is designed will
depend on the questions you want to answer. The two most familiar types of models
are visual models and physical models, but mathematical and computer models are
also common.
Visual models are two-dimensional representations of concepts, such as diagrams
and flow charts. Physical models are three-dimensional versions of reality that can
be scaled up or down.
Models help to make sense of ideas by visualising:
• objects that are difficult to see because of their size (too big or too small)
• processes that cannot easily be seen directly, such as feedback loops
• abstract ideas, such as energy transfer
• complex ideas, such as climate change.
CHAPTER 1 | WORKING SCIENTIFICALLY
1.6 Review
SUMMARY
• To make an investigation more meaningful, it
should be explained within the right context,
meaning the related physics ideas, concepts,
theories and models. Within this context, explain
the basis for the hypothesis.
• Models are useful tools that can be created and
used to gain a deeper understanding of concepts.
• Some common types of models are visual,
physical, mathematical and computational.
• Indicate whether the hypothesis is supported
or rejected and on what evidence this is based
(that is, the results). Do not provide irrelevant
information. Only refer to the specifics of the
hypothesis and the inquiry question and do not
make generalisations.
KEY QUESTIONS
1
Which of the following would not support a strong
conclusion to a report?
A The concluding paragraphs are relevant and provide
evidence.
B The concluding paragraphs are written in emotive
language.
C The concluding paragraphs include reference to
limitations of the research.
D The concluding paragraphs include suggestions for
further avenues of research.
2
Before you begin your investigation, you come up
with the hypothesis: According to Newton’s second
law, for a constant force, if the mass is increased the
acceleration is decreased.
After conducting the experiment you table your results
below:
Mass (kg)
Acceleration (ms−2)
1.0
3.0
2.0
2.0
3.0
1.0
3
You conduct an investigation to test the hypothesis: If
two objects are simultaneously dropped vertically from
the same height they will both land at the same time.
What is one conclusion you could reach if your results
found the following times for objects dropped from 1 m:
Object Time (s)
Feather
2
Tennis ball
0.5
Bowling ball
0.4
4
A procedure was repeated 30 times. How should the
following statement be rewritten?
Many repeats of the procedure were conducted.
If you were to analyse these results, how would they
support or refute your hypothesis?
CHAPTER 1 | WORKING SCIENTIFICALLY
35
1.7 Communicating
The way you approach communicating your results will depend on the audience
you want to reach. If you are communicating with a general audience you may want
to discuss your investigation in the style of a news article or blog post. These types of
communication don’t use too much scientific language as you need to assume that
your audience does not have a science background.
Throughout this course you will need to present your research using appropriate
nomenclature such as scientific language and notation. There are many different
presentation formats that you are used to such as posters, oral presentations and reports.
This section covers the main characteristics of effective science communication and
report writing, including objectivity, clarity, conciseness and coherence.
STRUCTURING A REPORT
Your report should have a clear, logical structure.
Introduction
•
SKILLBUILDER
Body paragraphs
L
Structuring body
paragraphs
The body paragraphs of a report
or essay need to be structured so
each idea is presented in a clear
way. Good paragraphs build up to
a report that has a logical flow.
One way to ensure each paragraph
is structured well is to use the
acronym TEEL: Topic, Elaborate,
Evidence, Link back.
Topic sentence
This establishes the key idea or
argument that will be put forward
in the paragraph. It supports the
main proposition of the overall
report.
Elaborate on the idea
•
•
•
•
•
•
•
•
The final section should summarise the main findings.
It should relate to the title of the investigation.
The conclusion should include limitations.
It should discuss implications and applications of the research and potential
future research.
Analysing information relevant to your
research investigation
Scientific research should always be objective and neutral. Any premise presented
must be supported with facts and evidence to allow the audience to make its own
decision. Identify the evidence supporting or contradicting each point you want to
make. It is also important to explain connections between ideas, concepts, theories
and models. Figure 1.7.1 lists the questions you need to consider when writing your
investigation report.
Do the sources
agree? What is
similar? What
is different?
Evidence
Provide evidence to support the
idea or argument in the topic
sentence.
Link back to the topic
sentence
Summarise the argument in the
paragraph and how it links to the
overall proposition set out by the
overall report.
Each subsequent paragraph should cover one main idea.
The first sentence of each paragraph should summarise the content of the
paragraph.
Use evidence to support statements.
Avoid very long or very short paragraphs.
Conclusion
Add further detail to the initial
topic sentence.
36
The first paragraph should introduce your research topic and define key terms.
Summarise main
ideas from sources.
Is the evidence
valid and reliable?
FIGURE 1.7.1
CHAPTER 1 | WORKING SCIENTIFICALLY
Have you identified
social, economic,
environmental and
ethical factors
relevant to your
research question?
Can you use a
labelled diagram
to help explain
concepts or
present evidence?
inquiry
question
What are the
limitations of the
information
you are
presenting?
What questions
do you have?
What are the
implications for
further research?
Questions you need to consider when writing your investigation report.
Once you have analysed your sources, annotate your outline, indicating where
you will use evidence and what the source of that evidence is. Try to introduce only
one idea per sentence and one theme per paragraph.
For example, for a report on ‘Experimental research into biodegradability of
plastics’, the third paragraph might contain information from:
• Selke et al. (2015), who reported no significant degradation
• Chiellini et al. (2007), who reported a significant degradation.
A report should include an analysis and synthesis of your sources.The information
from different sources needs to be explicitly connected, and it should be clear where
sources agree and disagree. In this example, the final sentences could be:
Selke et al. (2015) reported that tests of plastic polymers treated with
biodegradation additives resulted in no significant biodegradation after three years.
This finding contrasts with that of Chiellini et al. (2007), who reported significant
biodegradability of additive-treated polymers.
The different results can be explained by differences in the studies. The 2007
study tested degradation in natural river water, whereas the 2015 study tested
degradation under ultraviolet light, aerobic soil burial and anaerobic aqueous
conditions (Chiellini et al. 2007; Selke et al. 2015). As well as using different additives
and different experimental techniques, Selke et al. (2015) used additive rates of
1–5% and tested polyethylene terephthalate (PET) as well as polyethylene, whereas
Chiellini et al. (2007) used additive rates of 10–15% and tested only polyethylene.
Both studies were conducted under laboratory conditions, so they may not
reflect what happens in the natural environment.
WRITING FOR SCIENCE
Scientific reports are usually written in an objective or unbiased style. This is in
contrast with essay writing that often uses the subjective techniques of rhetoric or
persuasion. Read Table 1.7.1, which contrasts persuasive and scientific writing
styles.
TABLE 1.7.1
Persuasive writing versus scientific writing styles
Persuasive writing examples
Scientific writing equivalent examples
Use of biased and subjective language
Examples:
The results were extremely bad, atrocious,
wonderful etc.
This is terrible because …
This produced a disgusting odour.
Health crisis
Use of unbiased and objective language
Examples:
The results showed …
The implications of these results suggest …
The results imply …
This produced a pungent odour.
Health issue
Use of exaggeration
Example:
The object weighed a colossal amount, like
an elephant.
Use of non-emotive language
Example:
The object weighed 256 kg.
Use of everyday or colloquial language
Examples:
The bacteria passed away.
The results don’t …
The researchers had a sneaking
suspicion …
Use of formal language
Examples:
The bacteria died.
The results do not …
The researchers predicted/hypothesised/
theorised …
CHAPTER 1 | WORKING SCIENTIFICALLY
37
Consistent reporting narrative
Scientific writing can be written either in first-person or in third-person narrative.
Your teacher may advise you on which to select. In either case, ensure that you keep
the narrative point of view consistent. Read the examples of first-person and thirdperson narrative in Table 1.7.2.
TABLE 1.7.2
Examples of first-person and third-person narrative.
First person
Third person
I put 50 g of marble chips in a
conical flask and then added 10 mL
of 2 M hydrochloric acid.
First, 50 g of marble chips was weighed into the
conical flask and then 10 mL of 2 M hydrochloric
acid was added.
After I observed the reaction, I found
that…
After the reaction was completed, the results
showed…
My colleagues and I found…
Researchers found…
Qualified writing
Be careful of words that are absolute, such as always, never, shall, will and proven.
Sometimes it may be more accurate and appropriate to use qualifying words,
such as may, might, possible, probably, likely, suggests, indicates, appears, tends,
can and could.
Concise writing
It is important to write concisely, particularly if you want to engage and maintain
the interest of your audience. Use shorter sentences that are less verbose (contain
too many words). Read Table 1.7.3, which shows some examples of more concise
wording.
TABLE 1.7.3
Examples of verbose and concise language
Verbose example
Concise example
Due to the fact that
Because
Smith undertook an investigation into…
Smith investigated
It is possible that the cause could be…
The cause may be…
A total of five experiments
Five experiments
End result
Result
In the event that
If
Shorter in length
Shorter
Visual support
Identify concepts that can be explained using visual models and information that
can be presented in graphs or diagrams. This will not only reduce the word count of
your work but will also make it more accessible for your audience.
EDITING YOUR REPORT
Editing your report is an important part of the process. After editing your report, save
new drafts with a different file name and always back up your files in another location.
Pretend you are reading your report for the first time when editing. Once you
have completed a draft, it is a good idea to put it aside and return to it with ‘fresh
eyes’ a day later. This will help you find areas that need further work and give you
the opportunity to improve them. Look for content that is:
• ambiguous or unclear
• repetitive
38
CHAPTER 1 | WORKING SCIENTIFICALLY
•
•
•
•
•
•
awkwardly phrased
too lengthy
not relevant to your research question
poorly structured
lacking evidence
lacking a reference (if it is another researcher’s work).
Use a spellchecker tool to help you identify typographical errors, but first, check
that your spellchecker is set to Australian English. Also be wary of words that are
commonly misused, for example:
• where/were
• their/they’re/there
• affect/effect
• lead/led
• which/that.
REFERENCES AND ACKNOWLEDGEMENTS
All the quotations, documents, publications and ideas used in the investigation need
to be acknowledged in the ‘references and acknowledgments’ section in order to
avoid plagiarism and to ensure authors are credited for their work. References and
acknowledgements also give credibility to the study and allow the audience to locate
information sources should they wish to study it further.
When referencing a book, include in this order:
• author’s surname and initials; date of publication; title; publisher’s name; place
of publication.
For example:
Rickard G. et al. (2005), Science Dimensions 1, Pearson Education, Melbourne,
Victoria.
When referencing a website, include in this order:
• author’s surname and initials, or name of organisation, or title; year website was
written or last revised; title of webpage; date website was accessed; website address.
For example:
Wheeling Jesuit University/Center for Educational Technologies (2013),
NASA Physics Online Course: Forces and Motion, accessed 16 June 2015,
http://nasaphysics.cet.edu/forces-and-motion.html
MEASUREMENT AND UNITS
In every area of physics we have attempted to quantify the phenomena we study. In
practical demonstrations and investigations we generally make measurements and
process those measurements in order to come to some conclusions. Scientists have a
number of conventional ways of interpreting and analysing data from their investigations.
There are also conventional ways of writing numerical measurements and their units.
Correct use of unit symbols
The correct use of unit symbols removes ambiguity, as symbols are recognised
internationally. The symbols for units are not abbreviations and should not be
followed by a full stop unless they are at the end of a sentence.
Upper-case letters are not used for the names of any physical quantities of
units. For example, we write newton for the unit of force, while we write Newton
if referring to someone with that name. Upper-case letters are only used for the
symbols of the units that are named after people. For example, the unit of energy is
joule and the symbol is J. The joule was named after James Joule who was famous
for studies into energy conversions. The exception to this rule is ‘L’ for litre. We do
this because a lower-case ‘l’ looks like the numeral ‘1’. The unit of distance is metre
and the symbol is m. The metre is not named after a person.
CHAPTER 1 | WORKING SCIENTIFICALLY
39
The product of a number of units is shown by separating the symbol for each
unit with a dot or a space. Most teachers prefer a space but a dot is perfectly correct.
The division or ratio of two or more units can be shown in fraction form, using
a slash, or using negative indices. Most teachers prefer negative indices. Prefixes
should not be separated by a space. Table 1.7.4 shows some examples of the correct
use of units and symbols.
TABLE 1.7.4
Some examples of the use of symbols for derived units.
Incorrect
−2
ms
k Wh
−3
Preferred
−2
ms
kW h
Also correct
−
m.s 2, m/s2
kW.h
−3
−
kgm
kg m
kg.m 3, kg/m3
Nm
Nm
N.m
μm
μm
Units named after people can take the plural form by adding an ‘s’ when used
with numbers greater than one. Never do this with the unit symbols. It is acceptable
to say ‘two newtons’ but wrong to write 2 Ns. It is also acceptable to say ‘two newton’.
Numbers and symbols should not be mixed with words for units and numbers.
For example, twenty metres and 20 m are correct while 20 metres and twenty m are
incorrect.
Scientific notation
To overcome confusion or ambiguity, measurements are often written in scientific
notation. Quantities are written as a number between 1 and 10 and then multiplied
by an appropriate power of ten. Note that ‘scientific notation’, ‘standard notation’
and ‘standard form’ all have the same meaning.
Examples of some measurements written in scientific notation are:
0.054 m = 5.4 × 10−2 m
245.7 J = 2.457 × 102 J
2080 N = 2.080 × 103 N or 2.08 × 103 N
You should be routinely using scientific notation to express numbers. This
also involves learning to use your calculator intelligently. Scientific and graphics
calculators can be put into a mode whereby all numbers are displayed in scientific
notation. It is useful when doing calculations to use this mode rather than frequently
attempting to convert to scientific notation by counting digits on the calculator
display. It is quite acceptable to write all numbers in scientific notation, although
most people prefer not to use scientific notation when writing numbers between
0.1 and 1000.
An important reason for using scientific notation is that it removes ambiguity
about the precision of some measurements. For example, a measurement recorded
as 240 m could be a measurement to the nearest metre; that is, somewhere between
239.5 m and 240.5 m. It could also be a measurement to the nearest ten metres, that
is, somewhere between 235 m and 245 m. Writing the measurement as 240 m does
not indicate either case. If the measurement was taken to the nearest metre, it would
be written in scientific notation as 2.40 × 102 m. If it was taken to the nearest ten
metres only, it would be written as 2.4 × 102 m.
PREFIXES AND CONVERSION FACTORS
Conversion factors should be used carefully. You should be familiar with the
prefixes and conversion factors in Table 1.7.5. The most common mistake made
with conversion factors is multiplying rather than dividing. Some simple strategies
can save you this problem. Note that the table gives all conversions as a multiplying
factor.
40
CHAPTER 1 | WORKING SCIENTIFICALLY
Do not put spaces between prefixes and unit symbols. It is important to give the
symbol the correct case (upper or lower case). There is a big difference between
1 mm and 1 Mm.
TABLE 1.7.5
Prefixes and conversion factors.
Multiplying factor
Scientific notation
12
1 000 000 000 000
10
1 000 000 000
109
1 000 000
Prefix
Symbol
tera
T
giga
G
6
mega
M
3
kilo
k
−2
centi
c
−3
milli
m
−6
micro
μ
−9
nano
n
−12
pico
p
10
1 000
10
0.01
10
0.001
10
0.000 001
10
0.000 000 001
10
0.000 000 000 001
10
There is no space between prefixes and unit symbols. For example, onethousandth of an ampere is given the symbol mA. Writing it as m A is incorrect.
The space would make the symbol mean ‘metre ampere’.
FIGURE 1.7.2
A scientific calculator.
1.7 Review
SUMMARY
• A scientific report must include an introduction,
body paragraphs and conclusion.
• Scientific writing uses unbiased, objective,
accurate, formal language. Scientific writing
should also be concise and qualified.
• The conclusion should include a summary of the
main findings, a conclusion related to the issue
being investigated, limitations of the research,
implications and applications of the research, and
potential future research.
• Visual support can assist in conveying scientific
concepts and processes efficiently.
• Ensure you edit your final report.
• Scientific notation needs to be used when
communicating your results.
KEY QUESTIONS
1
Which of the following statements is written in
scientific style?
A The results were fantastic…
B The data in Table 2 indicates…
C The researchers felt…
D The smell was awful…
2
Which of the following statements is written in firstperson narrative?
A The researchers reported…
B Samples were analysed using…
C The experiment was repeated three times…
D I reported…
3
The variables acceleration, torque, momentum and
density each have different units. Write the units for
each of the following in correct scientific notation.
a acceleration; metres per second squared
b torque; newton metre
c momentum; kilogram metre per second
d density; kilogram per metre cubed
4
Convert 4.5 gigawatts (GW) into megawatts (MW).
5
Discuss why you might need to convert between
different multiplying factors, for example centimetres
to millimetres.
CHAPTER 1 | WORKING SCIENTIFICALLY
41
Chapter review
KEY TERMS
absolute error
accuracy
affiliation
bias
controlled variable
credible
dependent variable
expertise
hypothesis
independent variable
mean
median
mistake
mode
nomenclature
outlier
percentage uncertainty
personal protective
equipment (PPE)
persuasion
phenomenon
precision
primary source
qualitative variable
quantitative variable
random error
raw data
reliability
reputation
rhetoric
secondary source
significant figures
systematic error
trend
trend line
uncertainty
validity
variable
KEY QUESTIONS
1
What is a hypothesis and what form does it take?
2
The following steps of the scientific method are out of
order. Place a number (1–7) to the left of each point to
indicate the correct sequence.
Form a hypothesis.
Collect results.
Plan experiment and equipment.
Draw conclusions.
Question whether results support hypothesis.
State the inquiry question to be investigated.
Perform experiment.
3
42
List these types of hazard controls from the most
effective to the least effective:
substitution, personal protective equipment,
engineering controls, administrative controls,
elimination, isolation
4
Consider the hypothesis provided below. What are the
dependent, independent and controlled variables?
Hypothesis: Releasing an arrow in archery at an angle
greater or smaller than 45 degrees will result in a
shorter flight displacement (range).
5
What is the dependent variable in each prediction?
a If you push an object with a fixed mass with a larger
force, then the acceleration of that object will be
greater.
b The vertical acceleration of a falling object is
constant.
c A springboard diver rotates faster when in a tucked
position than when in a stretched (layout) position.
CHAPTER 1 | WORKING SCIENTIFICALLY
6
The speed of a toy car rolling down an inclined plane
was measured 6 times. The measurements obtained
(in cm s−1) were 7.0, 6.5, 6.8, 7.2, 6.5, 6.5. What is the
uncertainty and the mean of these values?
7
Which of the statistical measurements of mean, mode
and median is most affected by an outlier?
8
What relationship between variables is indicated by a
curved trend line?
9
If you hypothesise that impact force is directly
proportional to drop height, what would you expect a
graph of the data to look like?
10 What is meant by the ‘limitations’ of the investigation
method?
11 What is ‘bias’ in an investigation?
12 Which of the following is the correct way to reference
the source?
A Duffy et al. (2014) did find a dip in the star
formation rate.
B Duffy, A., Wyithe, S., Mutch, S. & Poole, G. (2014).
Low-mass galaxy formation and the ionizing photon
budget during reionization.
C Duffy, A., Wyithe, S., Mutch, S. & Poole, G. (2014).
Low-mass galaxy formation and the ionizing photon
budget during reionization, Monthly Notices of the
Royal Astronomical Society, 443(4), 3435–3443.
D Duffy et al. (2014) Low-mass galaxy formation and
the ionizing photon budget during reionization,
Monthly Notices of the Royal Astronomical Society.
13 Convert 2.5 mm (millimetres) into µm (micrometres).
14 List three things that need to be considered when
preparing a risk assessment.
15 A scientist designed and conducted an experiment
to test the following hypothesis: An increased
consumption of fast food causes a decrease in the
function of the liver.
a The discussion section of the scientist’s report
included comments on the accuracy, precision,
reliability, and validity of the investigation. Read
each of the following statements and determine
whether they relate to accuracy, reliability or
validity.
i Only teenage boys were tested.
ii Six boys were tested.
b The scientist then conducted the fast food study
with 50 people in the experimental group and 50
people in the control group. In the experimental
group, all 50 people gained weight. The scientist
concluded all the subjects gained weight as a result
of the experiment. Is this conclusion valid? Explain
why or why not.
c What recommendations would you make to the
scientist to improve the investigation?
16 What is the purpose of referencing and acknowledging
documents, ideas and quotations in your investigation?
17 You have measured the weight of an object using a set
of scales to be 200 g and the absolute uncertainty of
the scales is ±0.1 g. What is the percentage uncertainty
for this measurement?
20 Explain the meaning of the term ‘trend’ in a scientific
investigation and describe the types of trends that
might exist.
21 Explain the terms ‘accuracy’ and ‘validity’.
22 Which graph from the following list would be best to
use with each set of data listed here?
Graph types: pie diagram, scatter graph (with line of
best fit), bar graph, line graph
a The number of moons around each planet in the
solar system
b The temperature of water sampled at the same
time of day over a period of a month
c The magnitude of the gravitational constant at
different distances above sea level
d The proportion of energy being used by different
components in an electrical circuit
23 You are conducting an experiment that requires you
to measure the air temperature at the same time each
day. Using a thermometer which measures in units of
1°C you collect the following data.
Day
Temperature (°C)
1
2
3
4
5
22
24
21
25
27
a Calculate the absolute uncertainty in your data.
b What is the percentage uncertainty for the
temperature on day 3?
c Calculate the uncertainty from the mean.
18 a What is a controlled variable?
b What is a control experiment?
19 Identify which of the following pieces of information
about a cup of coffee are qualitative, and which are
quantitative. Place a tick in the appropriate column.
Information
Qualitative
Quantitative
cost $3.95
robust aroma
coffee temperature 82°C
cup height 9 cm
frothy appearance
volume 180 mL
strong taste
white cup
CHAPTER 1 | WORKING SCIENTIFICALLY
43
MODULE
Kinematics
Motion is a fundamental observable phenomenon. The study of kinematics
involves describing, measuring and analysing motion without considering the
forces and masses involved in that motion. Uniformly accelerated motion is
described in terms of relationships between measurable scalar and vector
quantities, including displacement, speed, velocity, acceleration and time.
Representations—including graphs and vectors, and equations of motion—can
be used qualitatively and quantitatively to describe and predict linear motion.
Outcomes
By the end of this module you will be able to:
• design and evaluate investigations in order to obtain primary and secondary
data and information PH11-2
• conduct investigations to collect valid and reliable primary and secondary data
and information PH11-3
• select and process appropriate qualitative and quantitative data and
information using a range of appropriate media PH11-4
• analyse and evaluate primary and secondary data and information PH11-5
• solve scientific problems using primary and secondary data, critical thinking
skills and scientific processes PH11-6
• describe and analyse motion in terms of scalar and vector quantities in
two dimensions and make quantitative measurements and calculations for
distance, displacement, speed, velocity and acceleration PH11-8.
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
CHAPTER
Motion in a straight line
Motion, from simple to complex, is a fundamental part of everyday life. In this
chapter you will learn how to use the mathematical quantities of scalars and vectors
to understand the concepts of forces and motion. From a train pulling in to a station
to a swimmer completing a lap of a pool, physics can model the motion of just
about anything.
Content
INQUIRY QUESTION
How is the motion of an object moving in a straight line
described and predicted?
By the end of this chapter you will be able to:
• describe uniform straight-line (rectilinear) motion and uniformly accelerated
motion through:
-- qualitative descriptions
-- the use of scalar and vector quantities (ACSPH060)
• conduct a practical investigation to gather data to facilitate the analysis of
instantaneous and average velocity through: ICT
-- quantitative, first-hand measurements
-- the graphical representation and interpretation of data (ACSPH061) N
• calculate the relative velocity of two objects moving along the same line using
vector analysis
• conduct practical investigations, selecting from a range of technologies, to record
and analyse the motion of objects in a variety of situations in one dimension in
order to measure or calculate: ICT N
-- time
-- distance
-- displacement
-- speed
-- velocity
-- acceleration
• use mathematical modelling and graphs, selected from a range of technologies,
to analyse and derive relationships between time, distance, displacement, speed,
velocity and acceleration in rectilinear motion, including:
2
-- s = ut + 21 at
-- v = u + at
-- v 2 = u2 + 2as (ACSPH061) ICT N
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
2.1 Scalars and vectors
WS
1.1
You will come into contact with many physical quantities in the natural world every
day. For example, time, mass and distance are all physical quantities. Each of these
physical quantities has units with which to measure them; for example, seconds,
kilograms and metres.
Some measurements only make sense if there is also a direction included. For
example, a GPS navigation system tells you when to turn and in which direction.
Without both of these two instructions, the information is incomplete.
All physical quantities can be divided into two broad groups based on what
information you need for the quantity to make sense. These groups are called
scalars and vectors. Often vectors are represented by arrows. Both of these types of
measures will be investigated throughout this section.
SCALARS
Many physical quantities can be described simply by a magnitude (size) and a
unit. For example, you might say that the speed of a car is 60 kilometres per hour.
The magnitude is 60 and the unit is kilometres per hour. Quantities like this that
have only a magnitude and a unit are called scalars.
The magnitude depends on the unit chosen. For example, 60 kilometres per
hour can also be described as 1 kilometre per minute. Here the magnitude is 1 and
the unit is kilometres per minute. Some examples of scalars are:
• time
• distance
• volume
• speed.
VECTORS
Sometimes a physical quantity also has a direction. For example, you might say that
a car is travelling at 60 kilometres per hour north. Quantities like this that have a
magnitude, a unit and a direction are called vectors. Some examples of vectors are:
• position
• displacement
• velocity
• acceleration
• force.
tail
head
10 m, east
head
tail
5 m, west
FIGURE 2.1.1 A simple vector diagram. The top
vector is twice as long as the bottom vector, so
it has twice the magnitude of the bottom vector.
The arrowheads indicate that the vectors are in
opposite directions.
48
MODULE 1 | KINEMATICS
Scalars are represented by a simple italic symbol, such as t for time and d for
distance.
Vectors are represented using vector notation. The most common type of vector
notation uses an arrow above the symbol. For example, force is written as F and
velocity is written as v . Without an arrow, F means only the magnitude of force
and v means only the magnitude of velocity.
You might see a different type of vector notation in books and journals. This uses
bold or bold italics to represent a vector, instead of an arrow. For example, force
is written as F or F and velocity is written as v or v.
VECTORS AS ARROWS
A vector has both a magnitude and a direction. Any vector can be represented
visually by an arrow. The length of the arrow represents the magnitude of the vector,
and the direction of the arrow (from tail to head) represents the direction of the
vector.
A diagram in which one or more vectors are represented by arrows is called a
vector diagram. Figure 2.1.1 is a vector diagram that shows two vectors.
A force is a push or a pull, and the unit of measure for force is the newton (N).
If you push a book to the right, it will respond differently compared to pushing it to
the left. So a force is described properly only when a direction is included, which
means that force is a vector. Forces are described in more detail in Chapter 4.
In most vector diagrams, the length of the arrow is drawn to scale so that it
accurately represents the magnitude of the vector.
In the scaled vector diagram in Figure 2.1.2, a force F = 4 N left acting on the toy
car is drawn as an arrow with a length of 2 cm. In this example a scale of 1 cm = 2 N
force is used.
An exact scale for the magnitude is not always needed, but it is important that
vectors are drawn accurately relative to one another. For example, a vector of 50 m
north should be half as long as a vector of 100 m south, and should point in the
opposite direction.
Point of application of arrows
→
F = 4 N left
FIGURE 2.1.2
toy car.
Vector diagrams may be presented slightly differently, depending on what they are
depicting. If the vector represents a force, the tail end of the arrow is placed at the
point where the force is applied to the object. If it is a displacement vector, the tail
is placed where the object started to move.
Figure 2.1.3 shows a force applied by a foot to a ball (95 N east) and an opposing
friction force (20 N west).
A force of 4 N left acts on a
E
W
DIRECTION CONVENTIONS
Vectors need a direction in order to make sense. However, there needs to be a way
of describing the direction that everyone understands and agrees upon.
Vectors in one dimension
20 N
For vector problems in one dimension, there are a number of direction conventions
that can be used. For example:
• forwards or backwards
• up or down
• left or right
• north or south
• east or west.
For vectors in one dimension there are only two possible directions. The two
directions must be in the same dimension or along the same line. The direction
convention that is used should be shown graphically in all vector diagrams. Some
examples are shown in Figure 2.1.4. Arrows like these are placed near the vector
diagram so that it is clear which convention is being used.
up
backwards
forwards
FIGURE 2.1.4
left
right
FIGURE 2.1.3 The force applied by the foot acts
at the point of contact between the ball and
the foot. The friction force acts at the point of
contact between the ball and the ground. The
kicking force, as indicated by the length of the
arrow, is much larger than the friction force.
N
W
down
95 N
E
S
Some common one-dimensional direction conventions.
Sign convention
In calculations involving one-dimensional vectors, a sign convention can also
be used to convert physical directions to the mathematical signs of positive and
negative. For example, forwards can be positive and backwards can be negative, or
right can be positive and left can be negative. A vector of 100 m up can be described
as −100 m, as long as the relationship between sign and direction conventions are
clearly indicated in a legend or key. Some examples are shown in Figure 2.1.4.
The advantage of using a sign convention is that the signs of positive and negative
can be entered into a calculator, while words such as ‘up’ and ‘right’ cannot. This is
useful when adding or subtracting vectors.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
49
Worked example 2.1.1
DESCRIBING VECTORS IN ONE DIMENSION
left
–
right
+
70 m
Describe the vector above using:
a the direction convention shown
Thinking
Working
Identify the magnitude and unit of the
vector.
The magnitude is 70 and the unit is
m (metres).
Identify the vector direction according
to the direction convention.
The vector is pointing to the right
according to the direction convention.
Combine the magnitude, unit and
direction.
The vector is 70 m right.
b the sign convention shown.
Thinking
Working
Convert the physical direction to the
corresponding mathematical sign.
The physical direction of right is
positive and left is negative. In this
example, the arrow is pointing right, so
the mathematical sign is +.
Combine the mathematical sign with
the magnitude and unit.
The vector is +70 m.
Worked example: Try yourself 2.1.1
DESCRIBING VECTORS IN ONE DIMENSION
west
–
east
+
50 N
Describe the vector above using:
a the direction convention shown
b the sign convention shown.
ADDING VECTORS IN ONE DIMENSION
Most real-life situations involve more than one vector acting on an object. If this is
the case, it is usually desirable to combine the vector diagrams to find the overall
effect of the vectors. When two or more vectors are in the same dimension, they are
said to be collinear (in line with each other).
This means that the vectors must either point in the same direction or point in
opposite directions. For example, the vectors 10 m west, 15 m east and 25 m west are
all in one dimension, because east is the opposite direction to west. When vectors are
combined, it is called adding vectors.
Graphical method of adding vectors
Vector diagrams, like those shown in Figure 2.1.5, are convenient for adding vectors.
To combine vectors in one dimension, draw the first vector, then start the second
vector with its tail at the head of the first vector. Continue adding arrows ‘head to
tail’ until the last vector is drawn. The sum of the vectors, or the resultant vector, is
drawn from the tail of the first vector to the head of the last vector.
50
MODULE 1 | KINEMATICS
s1 = 15 m east
→
s1 = 15 m east
s2 = 5 m east
→
→
+
=
s2 = 5 m east
→
sR = 20 m east
→
FIGURE 2.1.5 Adding vectors head to tail. This particular diagram represents the addition of 15 m
east and 5 m east. The resultant vector, shown in red, is 20 m east.
In Figure 2.1.5 the two vectors s1 (15 m east) and s2 (5 m east) are drawn
separately. The vectors are then redrawn with the head of s1 connected to the tail
of s2 . The resultant vector sR is drawn from the tail of s1 to the head of s2 . The
magnitude (size) of the resultant vector is the sum of the magnitudes of the separate
vectors: 15 m + 5 m = 20 m.
Alternatively, vectors can be drawn to scale; for example, 1 cm = 1 m. The
resultant vector is then measured directly from the scale diagram. The direction of
the resultant vector is the direction from the tail of the first vector to the head of the
last vector.
Algebraic method of adding vectors
To add vectors in one dimension using algebra, a sign convention is used to represent
the direction of the vectors, as in Figure 2.1.4 on page 49. When applying a sign
convention, it is important to provide a key explaining the convention used.
The sign convention allows you to enter the signs and magnitudes of vectors
into a calculator. The sign of the final magnitude gives the direction of the resultant
vector.
Vectors are added head to tail.
The resultant vector is drawn from
the tail of the first vector to the
head of the last vector.
Worked example 2.1.2
ADDING VECTORS IN ONE DIMENSION USING ALGEBRA
A student walks 25 m west, 16 m east, 44 m west and then 12 m east.
Use the sign convention in Figure 2.1.4 on page 49 to determine the resultant
displacement for the student.
Thinking
Working
Apply the sign convention
to change each of the
directions to signs.
25 m west = −25 m
16 m east = +16 m
44 m west = −44 m
12 m east = +12 m
Add the magnitudes and
their signs together.
Resultant vector = (−25) + (+16) + (−44) + (+12)
Refer to the sign and
direction conventions to
determine the direction of
the resultant vector.
Negative is west.
State the resultant vector.
The resultant vector is 41 m west.
= −41 m
Worked example: Try yourself 2.1.2
ADDING VECTORS IN ONE DIMENSION USING ALGEBRA
A box has the following forces acting on it: 16 N up, 22 N down, 4 N up and 17 N
down. Use the sign and direction conventions in Figure 2.1.4 on page 49 to
determine the resultant force on the box.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
51
PHYSICSFILE
SUBTRACTING VECTORS IN ONE DIMENSION
N
Double negatives
It is important to differentiate between
the terms subtract, minus, take away
or difference between and the term
negative. The terms subtract, minus,
take away or difference between are
processes, like add, multiply and divide.
You will find them grouped together
on your calculator. The term negative
is a property of a number that means
that it is opposite to positive. There is a
separate button on your calculator for
this property.
When a negative number is subtracted
from a positive number, the two numbers
are added together. For example,
5 − (−2) = 7.
To find the difference between two vectors, you must subtract the initial vector
from the final vector. To do this, work out which is the initial vector, then reverse its
direction to obtain the opposite of the initial vector. Then add the final vector to the
opposite of the initial vector.
This technique can be applied both graphically and algebraically.
Graphical method of subtracting vectors
Velocity indicates how fast an object is moving, and in what direction. It is a vector
because it involves both magnitude and direction. For example, in Figure 2.1.6 the
velocity of the tennis ball as it hits the racquet is different from the velocity of the
ball when it leaves the racquet, because the ball has changed direction. The concept
of velocity is covered in more detail in Section 2.2, but it is useful to use the example
of velocity now when discussing the subtraction of vectors. The processes applied to
the subtraction of velocity vectors works for all other vectors.
To subtract velocity vectors in one dimension using a graphical method,
determine which vector is the initial velocity and which is the final velocity. The
final velocity is drawn first. The initial velocity is then drawn, but in the opposite
direction to its original form. The sum of these vectors, or the resultant vector, is
drawn from the tail of the final velocity to the head of the reversed initial velocity.
This resultant vector is the difference between the two velocities, ∆v .
The mathematical symbol Δ (delta) is used to represent the change in a
variable. For example, ∆v means the change in velocity.
To find the difference between two vectors in the same dimension, subtract the
initial vector from the final vector. Vectors are subtracted by adding the negative
of one vector to the positive of the other vector.
Velocity is a vector, so its direction
is important. The velocity of a tennis ball
immediately before it hits a racquet is different
to its velocity immediately after it leaves the
racquet, because it is travelling in a different
direction.
FIGURE 2.1.6
In Figure 2.1.7, two velocity vectors v1 (9 m s−1 east) and v2 (3 m s−1, east) are
drawn separately. The initial velocity v1 is then redrawn in the opposite direction to
form −v1 or 9 m s−1 west.
v2 = 3 m s–1 east
→
v1 = 9 m s–1 east
→
FIGURE 2.1.7
∆v = 6 m s–1 west
→
v2 = 3 m s–1 east
→
–v1 = 9 m s–1 west
→
FIGURE 2.1.8 Subtracting vectors using the
graphical method.
–v1 = 9 m s–1 west
→
Subtracting vectors using the graphical method
Figure 2.1.8 illustrates how the difference between the vectors is found. First
the final velocity, v2 , is drawn. Then the opposite of the initial velocity, −v1, is drawn
head to tail. The resultant vector, ∆v , is drawn from the tail of v2 to the head of −v1.
The magnitude of the resultant vector, ∆v , can be calculated from the magnitudes
of the two vectors. Alternatively, you could draw the vectors to scale and then
measure the resultant vector against that scale, for example 1m s−1 = 1 cm.
The direction of the resultant vector, ∆v, is the same as the direction from the tail
of the final velocity, v2 , to the head of the opposite of the initial velocity, −v1.
Algebraic method of subtracting vectors
To subtract velocity vectors in one dimension algebraically, a sign convention is
used to represent the direction of the velocities. Some examples of one-dimensional
directions include east and west, north and south and up and down. These options
are replaced by positive (+) or negative (–) signs when calculations are performed.
To change the direction of the initial velocity, simply change the sign from positive
to negative or from negative to positive.
52
MODULE 1 | KINEMATICS
The equation for finding the change in velocity is:
change in velocity = final velocity − initial velocity
∆v = v2 − v1
which is the same as:
change in velocity = final velocity + the opposite of the initial velocity
∆v = v2 + (−v1 )
The final velocity is added to the opposite of the initial velocity. Because the change
in velocity is a vector, it will consist of a sign, a magnitude and a unit. The sign of the
answer can be compared with the sign and direction convention (Figure 2.1.9) to
determine the direction of the change in velocity.
Worked example 2.1.3
up +
down –
backwards
v1 = 255 m s−1 west = −255 m s−1
v2 = 160 m s−1 east = +160 m s−1
Reverse the direction of the initial
velocity v1 by reversing the sign.
−v1 = 255 m s−1 east
Use the formula for change in velocity
to calculate the magnitude and the
sign of ∆v.
∆v = v2 + (−v1 ) = (+160) + (+255)
Refer to the sign and direction
convention to determine the direction
of the change in velocity.
Positive is east.
State the resultant vector.
The resultant vector is 415 m s−1 east.
+
N+
S–
W
–
Apply the sign and direction
convention to change the directions
to signs.
right
–
An aeroplane changes course from 255 m s−1 west to 160 m s−1 east. Use the sign
and direction conventions in Figure 2.1.9 to determine the change in the velocity
of the aeroplane.
Working
+
left
SUBTRACTING VECTORS IN ONE DIMENSION USING ALGEBRA
Thinking
forwards
–
E
+
FIGURE 2.1.9 One-dimensional direction
conventions can also be expressed as sign
conventions.
= +255 m s−1
= +415 m s−1
Worked example: Try yourself 2.1.3
SUBTRACTING VECTORS IN ONE DIMENSION USING ALGEBRA
A rocket accelerates from 212 m s−1 to 2200 m s−1 upwards. Use the sign and
direction conventions in Figure 2.1.9 to determine the change in velocity of the
rocket.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
53
2.1 Review
SUMMARY
• Scalar quantities have only a magnitude and a
unit.
• Adding vectors in one dimension can be done
algebraically by applying a sign convention.
Vectors with direction become vectors with either
positive or negative signs.
• Vector quantities have a magnitude, a unit and
a direction. They are represented using vector
notation. An arrow above the variable indicates
that it is a vector.
• To find the difference between two vectors,
subtract the initial vector from the final vector.
• Combining vectors is known as adding vectors.
• Vectors are subtracted by adding the negative, or
opposite, of a vector.
• Adding vectors in one dimension can be done
graphically using vector diagrams. After adding
vectors head to tail, the resultant vector can be
drawn from the tail of the first vector to the head
of the last vector.
• Subtracting vectors in one dimension can be done
graphically using a scale.
• Subtracting vectors in one dimension can also be
done algebraically.
KEY QUESTIONS
Add the following vectors to find the resultant vector:
3 m up, 2 m down and 3 m down.
A car that was initially travelling at a velocity of 3 m s−1
west is later travelling at 5 m s−1 east. What is the
difference between the two vectors?
6
Determine the resultant vector of a model train that
moves in these directions: 23 m forwards, 16 m
backwards, 7 m forwards and 3 m backwards.
Determine the change in velocity of a runner who
changes from running at 4 m s−1 to the right on grass
to running 2 m s−1 to the right in sand.
7
A student throws a ball up into the air at 4 m s−1. A
short time later the ball is travelling back downwards
to hit the ground at 3 m s−1. Determine the change in
velocity of the ball during this time.
8
Jamelia applies the brakes on her car and changes
her velocity from 22.2 m s−1 forwards to 8.2 m s−1
forwards. Calculate the change in velocity of
Jamelia’s car.
Find the resultant vector when the following vectors
are combined: 2 m west, 5 m east and 7 m west.
2
3
4
54
5
1
When adding vector B to vector A using the head to
tail method, from what point, and to what point, is the
resultant vector drawn?
A from the head of A, to the tail of B
B from the tail of B, to the head of A
C from the head of B, to the tail of A
D from the tail of A, to the head of B
MODULE 1 | KINEMATICS
2.2 Displacement, speed and velocity
In order to describe and analyse motion, it is important to understand the terms
used to describe it, even in its simplest form. In this section you will learn about
some of the terms used to describe rectilinear or straight-line motion, such as
position, distance, displacement, speed and velocity.
(a)
CENTRE OF MASS
Motion is often more complicated than it seems at first. For example, when a
freestyle swimmer travels at a constant speed of 2 m s−1, their head and torso move
forwards at this speed, but the motion of their arms is more complex. At times their
arms move forwards through the air faster than 2 m s−1, and at other times they
move backwards through the water.
It is beyond the scope of this course to analyse such a complex motion. However,
the motion of the swimmer can be simplified by treating the swimmer as a simple
object located at a single point called the centre of mass. The centre of mass is the
balance point of an object. For a person, the centre of mass is just above the waist.
The centres of mass of some objects are shown in Figure 2.2.1.
(b)
POSITION, DISTANCE AND DISPLACEMENT
Position
One important term to understand when analysing straight-line motion is position.
• Position describes the location of an object at a certain point in time with respect
to the origin.
• Position is a vector quantity and therefore requires a direction.
Consider Sophie doing laps in a 50 m pool, as shown in Figure 2.2.2. To simplify
her motion, Sophie is treated as a point object with her centre of mass just above her
waist. The pool can be treated as a one-dimensional number line, with the starting
block defined as the origin. The direction to the right of the starting block is taken
to be positive.
Sophie’s position as she is warming up behind the starting block is −10 m, as
shown in Figure 2.2.2(a). The negative sign indicates the direction from the origin,
i.e. to the left. Her position could also be described as 10 m to the left of the starting
block.
At the starting block, Sophie’s position is 0 m, as shown in Figure 2.2.2(b). Then
after swimming a little over half a length of the pool she is +25 m or 25 m to the right
of the origin, as shown in Figure 2.2.2(c).
(a)
–10
0
10
20
30
40
50
–10
0
10
20
30
40
50
–10
0
10
20
30
40
50
(c)
(d)
FIGURE 2.2.1 The centre of mass of each object
is indicated by a cross.
60 m
Position
(b)
60 m
Position
(c)
60 m
Position
FIGURE 2.2.2 The position of the swimmer is given with reference to the starting block. (a) While
warming up, Sophie is at −10 m. (b) When she is on the starting block, her position is zero. (c) After
swimming for a short time, she is at a position of +25 m.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
55
Distance travelled
Position describes where an object is at a certain point in time. But distance
travelled is how far a body travels during a journey. For example, the tripmeter
or odometer of a car or bike measures distance travelled. Distance travelled is
represented by the symbol d.
• Distance travelled (d) describes the length of the path covered during an
object’s entire journey.
• Distance travelled is a scalar quantity and is measured in metres (m).
For example, if Sophie completes three lengths of the pool, the distance travelled
during her swim will be 50 + 50 + 50 = 150 m.
The distance travelled is not affected by the direction of the motion. That is, the
distance travelled by an object always increases as it moves, regardless of its direction.
Displacement
Displacement is the change in position of an object, and is represented by the
symbol s . Displacement considers only where the motion starts and finishes. The
route taken between the start and finish has no effect on displacement. If the motion
is in one dimension, a word such as east or west, or a positive or negative sign, can
be used to indicate the direction of the displacement.
• Displacement is
the change in position of an object in a given direction.
• Displacement s = final position − initial position.
• Displacement is a vector quantity and is measured in metres (m).
Consider the example of Sophie completing one length of the pool. During
her swim, the distance travelled is 50 m. Her final position is +50 m and her initial
position is 0 m. Her displacement is:
s = final position − initial position
= 50 − 0
= +50 m or 50 m in a positive direction
Notice that magnitude, units and direction are required for a vector quantity. The
distance will be equal to the magnitude of displacement only if the body is moving
in a straight line and does not change direction. If Sophie swims two lengths, her
distance travelled will be 100 m (50 m out and 50 m back) but her displacement
during this swim will be:
s = final position − initial position
=0−0
= 0m
Even though Sophie has swum 100 m, her displacement is zero because the
initial and final positions are the same.
The above formula for displacement is useful if you already know the initial and
final positions of the object. An alternative method to determine total displacement,
if you know the displacement of each section of the motion, is to add up the
individual displacements for each section of motion.
The total displacement is the sum of individual displacements.
GO TO ➤ Section 2.1, page 50
56
MODULE 1 | KINEMATICS
It is important to remember that displacement is a vector and so, when adding
displacements, you must obey the rules of vector addition (discussed in Section 2.1).
In the example above, in which Sophie completed two laps, overall displacement
could have been calculated by adding the displacement of each lap:
s = sum of displacements for each lap
= 50 m in the positive direction + 50 m in the negative direction
= 50 + (−50)
= 0m
SPEED AND VELOCITY
For thousands of years, humans have tried to travel at ever greater speeds. This
desire has contributed to the development of all sorts of competitive activities, as
well as major advances in engineering and design. World records for some of these
pursuits are given in Table 2.2.1.
TABLE 2.2.1
World record speeds for a variety of sports or modes of transport
World record speed (m s−1)
World record speed (km h−1)
luge
43
140
train
167.5
603
tennis serve
73.1
263
waterskiing (barefoot)
60.7
218
cricket delivery
44.7
161
racehorse
19.7
71
Activity or object
Speed and velocity are both quantities that give an indication of how quickly
the position of an object is changing. Both terms are in common use and are often
assumed to have the same meaning. In physics, however, these two terms have
different definitions.
• Speed is the rate at which distance is travelled. Like distance, speed is a
scalar. A direction is not required when stating the speed of an object.
• Velocity is the rate at which displacement changes. It has direction, so it is a
vector quantity. A direction must always be given when stating the velocity of
an object.
• The standard SI unit for speed and velocity is metres per second (m s−1).
Instantaneous speed and velocity
Instantaneous speed and instantaneous velocity tell you how fast something is
moving at a particular point in time. The speedometer on a car or bike indicates
instantaneous speed.
If a speeding car is travelling north and is detected on a police radar gun at
150 km h−1, it indicates that this car’s instantaneous speed is 150 km h−1, while its
instantaneous velocity is 150 km h−1 north. The instantaneous speed is always equal
to the magnitude of the instantaneous velocity.
Average speed and velocity
Average speed and average velocity both give an indication of how fast an object is
moving over a particular time interval.
travelled = d
average speed v av = distance
∆t
time taken
s
average velocity v av = displacement
time taken = ∆t
CHAPTER 2 | MOTION IN A STRAIGHT LINE
57
Average speed is equal to instantaneous speed only when a body’s motion is
uniform; that is, if it is moving at a constant speed.
The average speed of a car that takes 30 minutes to travel 20 km from Macquarie
Park to Manly Beach is 40 km h−1. But this does not mean that the car travelled the
whole distance at this speed. In fact it is more likely that the car was sometimes
moving at 60 km h−1, and at other times was not moving at all.
A direction (such as north, south, up, down, left, right, positive, negative) must
be given when describing a velocity. The direction of velocity is always the same
as the direction of displacement. And like the relationship between distance and
displacement, average speed will be equal to the magnitude of average velocity only
if the body is moving in a straight line and does not change direction.
For example, in a cycling race of one lap around a track (Figure 2.2.3), the
magnitude of the average velocity will be zero, because the displacement is zero.
This is true no matter what the average speed for the lap is.
SKILLBUILDER
N
Converting units
Australian Anna Meares won
the UCI Mexico Track World Cup 2013. She
rode 500 m in a world record time of 32.836 s.
Her average speed around the track was
55.6 km h−1, but her average velocity was zero.
FIGURE 2.2.3
PHYSICSFILE N
Reaction time
Drivers are often distracted by loud
music or phone calls. These distractions
result in many accidents and deaths on
the road. If cars are travelling at high
speeds, they will travel a large distance
in the time that the driver takes just to
apply the brakes. A short reaction time
is very important for all road users.
This is easy to understand given the
relationship between speed, distance
and time.
distance travelled = v × t
The usual unit used in physics for velocity is m s−1, but km h−1 is often used in
everyday life. So it is important to understand how to convert between these
two units.
Converting km h−1 to m s−1
You should be familiar with 100 km h−1 because it is the speed limit for
most freeways and country roads in Australia. Cars that maintain this speed
would travel 100 km in 1 hour. Since there are 1000 metres in 1 kilometre
and 3600 seconds in 1 hour (60 s × 60 min), this is the same as travelling
100 000 m in 3600 s.
100kmh–1 = 100 × 1000mh–1
= 100 000mh–1
000
= 100
ms−1
3600
= 27.8ms –1
So km h−1 can be converted to m s−1 by multiplying by
Converting
m s−1
to
km h−1
1000
3600
(or dividing by 3.6).
A champion Olympic sprinter can run at an average speed of close to 10 m s−1.
Each second, the athlete will travel approximately 10 metres. If they could
maintain this rate, in 1 hour the athlete would travel 10 × 3600 = 36 000 m =
36 km.
10ms –1 = 10 × 3600mh–1
= 36 000mh–1
=
36 000
kmh−1
1000
= 36kmh–1
So m s−1 can be converted to km h−1 by
3600 or 3.6.
multiplying by 1000
÷ 3.6
When converting a speed from one unit
km h–1
m s–1
to another, it is important to think about
the speeds to ensure that your answers
× 3.6
make sense. The diagram in Figure 2.2.4
converting
summarises the conversion between the FIGURE 2.2.4−1Rules for −1
between m s and km h .
two common units for speed.
58
MODULE 1 | KINEMATICS
Worked example 2.2.1
AVERAGE VELOCITY AND CONVERTING UNITS
Sam is an athlete performing a training routine by running back and forth along a
straight stretch of running track. He jogs 100 m north in a time of 20 s, then turns
and walks 50 m south in a further 25 s before stopping.
a What is Sam’s average velocity in m s−1?
Thinking
Working
Calculate the displacement.
(Remember that total displacement is
the sum of individual displacements.)
s = sum of displacements
Sam’s total journey consists of two
displacements: 100 m north and then
50 m south.
= 100 m north + 50 m south
= 100 + (−50)
= +50 m or 50 m north
start
N 0
S
20
finish
100 m
start
45
0
finish
50 m
Work out the total time taken for the
journey.
20 + 25 = 45 s
Substitute the values into the velocity
equation.
Displacement is 50 m north.
Time taken is 45 s.
Average velocity vav =
=
s
∆t
50
45
= 1.1ms−1
Velocity is a vector, so a direction must
be given.
1.1 m s−1 north
b What is the magnitude of Sam’s average velocity in km h−1?
Thinking
Working
Convert from m s−1 to km h−1 by
multiplying by 3.6.
vav = 1.1m s –1
= 1.1 × 3.6
= 4.0 kmh–1 north
As the magnitude of velocity is needed,
direction is not required in this answer.
v av = 4.0 km h−1
CHAPTER 2 | MOTION IN A STRAIGHT LINE
59
c What is Sam’s average speed in m s−1?
Thinking
Working
Calculate the distance. (Remember
that distance is the length of the path
covered over the entire journey. The
direction does not matter.)
d = 100 + 50
= 150
Sam travels 100 m in one direction
and then 50 m in the other direction.
Work out the total time taken for the
journey.
20 + 25 = 45 s
Substitute the values into the speed
equation.
Distance is 150 m.
Time taken is 45 s.
Average speed vav =
d
∆t
= 150
45
= 3.3 ms−1
d What is Sam’s average speed in km h−1?
Thinking
from m s−1
Convert
multiplying by 3.6.
Working
to km h−1
by
Average speed vav = 3.3 m s−1
= 3.3 × 3.6
= 12 km h−1
Worked example: Try yourself 2.2.1
AVERAGE VELOCITY AND CONVERTING UNITS
Sally is an athlete performing a training routine by running backwards and forwards
along a straight stretch of running track. She jogs 100 m west in a time of 20 s, then
turns and walks 160 m east in a further 45 s before stopping.
a What is Sally’s average velocity in m s−1?
b What is the magnitude of Sally’s average velocity in km h−1?
c What is Sally’s average speed in m s−1?
d What is Sally’s average speed in km h−1? Give your answer to two significant
figures.
60
MODULE 1 | KINEMATICS
PHYSICS IN ACTION
N
Alternative units for speed and distance
Metres per second is the standard unit for measuring
speed because it is derived from the standard unit
for distance (metres) and the standard unit for time
(seconds). However, alternative units are often used to
better suit a certain application.
The speed of a boat is usually measured in knots,
where 1 knot = 0.51 m s−1. This unit originated in the
nineteenth century, when the speed of sailing ships would
be measured by allowing a rope, with knots tied at regular
intervals, to be dragged by the water through a sailor’s
hands. By counting the number of knots that passed
through the sailor’s hands, and measuring the time taken
for this to happen, the average speed formula could be
applied to estimate the speed of the ship.
The speed of very fast aircraft, such as the one in
Figure 2.2.5, is often stated using Mach numbers. A speed
of Mach 1 equals the speed of sound, which is 340 m s−1 at
the Earth’s surface. Mach 2 is twice the speed of sound, or
680 m s−1, and so on.
The light-year is an alternative unit for measuring distance.
The speed of light in a vacuum is nearly 300 000 km s−1.
FIGURE 2.2.5 Modern fighter aircraft are able to fly at speeds well
above Mach 2.
One light-year is the distance that light travels in one year.
Astronomers use this unit because distances between
objects in the universe are enormous. It takes about 4.24
years for light to travel from the nearest star (Proxima
Centauri) to us. That means the distance from our solar
system to the nearest star is about 4.24 light-years. Light
takes about 8.5 minutes to travel from the Sun to Earth, so
you could say that the Sun is 8.5 light-minutes away.
2.2 Review
SUMMARY
• The motion of an object travelling in a straight line
is called rectilinear motion.
• Position defines the location of an object with
respect to a defined origin.
• Distance travelled, d, tells us how far an object has
actually travelled. Distance travelled is a scalar.
• Displacement, s, is a vector and is defined as
the change in position of an object in a given
direction: s = final position − initial position.
• The average speed of a body, vav, is defined as the
rate of change of distance and is a scalar quantity:
average speed v av =
distance travelled
time taken
=
• The average velocity of a body, v av, is defined as
the rate of change of displacement and is a vector
quantity:
average velocity v av = displacement
= ∆st
time taken
• The usual SI unit for both speed and velocity is
metres per second (m s−1); kilometres per hour
(km h−1) is another SI unit that is commonly used.
• To convert from m s−1 to km h−1, multiply by 3.6.
• To convert from km h−1 to m s−1, divide by 3.6.
d
∆t
CHAPTER 2 | MOTION IN A STRAIGHT LINE
61
2.2 Review continued
KEY QUESTIONS
1
A girl swims 10 lengths of a 25 m pool. Which one or
more of the following statements correctly describes
her distance travelled and displacement?
A Her distance travelled is zero.
B Her displacement is zero.
C Her distance travelled is 250 m.
D Her displacement is 250 m.
2
An insect is walking back and forth along a metre
ruler, as show in the figure below. Taking the right as
positive, determine both the size of the displacement
and the distance travelled by the insect as it travels on
the following paths.
A
0
10
a
b
c
d
20
30
B
C
40
50
D
60
70
80
90
100
cm
A to B
C to B
C to D
C to E and then to D.
3
During a training ride, a cyclist rides 50 km north, then
30 km south.
a What is the distance travelled by the cyclist during
the ride?
b What is the displacement of the cyclist for this ride?
4
A lift in a city building, shown in the figure below,
carries a passenger from the ground floor down to the
basement, then up to the top floor.
a What is the
displacement of
the lift as it travels
from the ground
top floor
floor to the
basement?
b What is the
displacement
of the lift as it
travels from the
basement to the
50 m
top floor?
c What is the
distance travelled
by the lift during
this entire trip?
d What is the
ground
displacement of
floor
the lift during this
entire trip?
10 m
MODULE 1 | KINEMATICS
A car travelling at a constant speed was timed over
400 m and was found to cover the distance in 12 s.
a What was the car’s average speed?
b The driver was distracted and his reaction time was
0.75 s before applying the brakes. How far did the
car travel in this time?
6
A cyclist travels 25 km in 90 minutes.
a What is her average speed in km h−1?
b What is her average speed in m s−1?
7
Liam pushes his toy truck 5 m east, then stops it
and pushes it 4 m west. The entire motion takes
10 seconds.
a What is the truck’s average speed?
b What is the truck’s average velocity?
8
An athlete in training for a marathon runs 10 km north
along a straight road before realising that she has
dropped her drink bottle. She turns around and runs
back 3 km to find her bottle, then resumes running
in the original direction. After running for 1.5 h, the
athlete stops when she is 15 km from her starting
position.
a What is the distance travelled by the athlete during
the run?
b What is the athlete’s displacement during the run?
c What is the average speed of the athlete in km h−1?
d What is the athlete’s average velocity in km h−1?
E
basement
62
5
2.3 Acceleration
PHYSICS INQUIRY
CCT N
Modelling acceleration
How is the motion of an object moving in a straight line
described and predicted?
COLLECT THIS…
•
5 large metal nuts
•
5 m length of string
•
metal baking tray
•
ruler or measuring tape
DO THIS…
5 Drop the string and listen to the
beats the nuts make as they hit
the tray.
6 Using trial and error, adjust the
position of the nuts on the string so
they create equally spaced beats
when they hit the baking tray.
1 Tie one nut onto the end of the
string.
RECORD THIS…
2 Thread the other nuts onto
the string one at a time, tying
the string to fix them at 10 cm
intervals.
Describe the spacing between each
nut that produced equally spaced
beats.
3 Place the metal baking tray on the
ground, upside down.
4 Standing above the tray, hold the
string so that the first nut is just
resting on the tray.
Present your results in a table.
REFLECT ON THIS…
How is the motion of an object
moving in a straight line described
and predicted?
Explain why the pattern observed in
this experiment was created.
Acceleration is a measure of how quickly velocity changes. When you are in a car
that speeds up or slows down, you experience acceleration. In an aircraft taking
off along a runway, you experience a much greater acceleration. Because velocity
has magnitude and direction, acceleration can be caused by a change in speed or a
change in direction. In this section you will look at the simple case of acceleration
caused by a change in velocity while travelling in a straight line.
FINDING THE CHANGE IN VELOCITY AND SPEED
The velocity and speed of everyday objects are changing all the time. Examples of
these are when a car moves away as the traffic lights turn green, when a tennis ball
bounces or when you travel on a rollercoaster.
If the initial and final velocity of an object are known, its change in velocity
can be calculated. To find the change in any physical quantity, including speed and
velocity, the initial value is subtracted from the final value.
Vector subtraction was covered in detail in Section 2.1.
Change in speed is the final
speed minus the initial speed:
Δv = v − u
where u is the initial speed in m s−1
v is the final speed in m s−1
Δv is the change in speed
in m s−1.
Since speed is a scalar, direction
is not required.
GO TO ➤ Section 2.1, page 52
Change in velocity is the final velocity minus the initial velocity:
∆v = v − u
where u is the initial velocity in m s−1
v is the final velocity in m s−1
∆v is the change in velocity in m s−1.
Because velocity is a vector, this should be done using vector subtraction. Like all
vectors, velocity must include a direction.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
63
Worked example 2.3.1
CHANGE IN SPEED AND VELOCITY PART 1
A ball is dropped onto a concrete floor and strikes the floor at 5.0 m s−1. It then
rebounds at 5.0 m s−1.


u = 5 m s–1
v = 5 m s–1
a What is the change in speed of the ball?
Thinking
Working
Find the values for the initial speed
and the final speed of the ball.
u = 5.0 m s−1
Substitute the values into the change
in speed equation: Δv = v − u
Δv = v − u
v = 5.0 m s−1
= (5.0) − (5.0)
= 0 m s−1
b What is the change in velocity of the ball?
Thinking
Velocity is a vector. Apply the sign
convention to replace the directions.
Working
u = 5.0 m s−1 down
= −5.0 m s−1
v = 5.0 m s−1 up
= +5.0 m s−1
As this is a vector subtraction, reverse
the direction of u to get −u .
Substitute the values into the change
in velocity equation:
∆v = v + (−u )
Apply the sign convention to describe
the direction.
u = −5.0 m s−1, therefore
−u = +5.0 m s−1
∆v = v + (−u )
= (+5.0) + (+5.0)
= +10 m s−1
∆v = 10.0 m s−1 up
Worked example: Try yourself 2.3.1
CHANGE IN SPEED AND VELOCITY PART 1
A ball is dropped onto a concrete floor and strikes the floor at 9.0 m s−1. It then
rebounds at 7.0 m s−1.
a What is the change in speed of the ball?
b What is the change in velocity of the ball?
64
MODULE 1 | KINEMATICS
ACCELERATION
Consider the following information about the velocity of a car that starts from rest,
as shown in Figure 2.3.1. The velocity of the car increases by 10 km h−1 to the right
each second. If right is taken to be the positive direction, the car’s velocity changes
by +10 km h−1 per second, or +10 km h−1 s−1.
t=0s
FIGURE 2.3.1
1s
2s
A car’s acceleration as its velocity increased from
0 km h−1
to
+30 km h−1.
3s
The athlete in Figure 2.3.2 takes 3 seconds to come to a stop at the end of a race.
The velocity of the athlete changes by −2 m s−1 to the right each second. If right is
taken to be the positive direction, the athlete’s acceleration is −2 metres per second
per second, or −2 m s−2.
v = 6 m s–1
4 m s–1
2 m s–1
0 m s–1
t=0s
1s
2s
3s
FIGURE 2.3.2
−2 m s−2.
The velocity of the athlete changes by
−2 m s−1
each second. The acceleration is
When the direction of motion is taken to be the positive direction, a negative
acceleration means that the object is slowing down in the direction of travel, like the
athlete in Figure 2.3.2. A negative acceleration can also mean the object is speeding
up but in the opposite direction.
Because acceleration is a vector quantity, a vector diagram can be used to find
the resultant acceleration of an object.
Average acceleration
Like speed and velocity, the average acceleration of an object can also be calculated.
To do this you need to known how long the change in velocity lasted.
Average acceleration, aav, is the rate of change of velocity:
in velocity
aav = change
change in time
= ∆∆vt
= v ∆−tu
where v is the final velocity in m s−1
u is the initial velocity in m s−1
Δt is the time interval in seconds.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
65
PHYSICS IN ACTION
ICT
Human acceleration
In the 1950s the United States Air Force used rocket
sleds (Figure 2.3.3) to study the effects of extremely large
accelerations on humans, with the aim of improving the
chances of pilots surviving crashes. At that time it was
thought that humans could not survive accelerations above
about 175 m s−2, so aircraft seats, harnesses and cockpits
were not designed to withstand larger accelerations.
FIGURE 2.3.3 A rocket-powered sled used to test the effects of
acceleration on humans.
One volunteer, Colonel John Stapp, was strapped into a
sled and accelerated to speeds of over 1000 km h−1 in a
very short time. Water scoops were used to stop the sled
in less than 2 seconds, producing a deceleration of more
than 450 m s−2. The effects of these massive accelerations
are evident on his face (Figure 2.3.4).
The results showed that humans could survive much
higher decelerations than previously thought, and that
the mechanical failure of seats, harnesses and cockpit
structures were major causes of deaths in aircraft
accidents.
FIGURE 2.3.4 Photos showing the distorted face of Colonel John Stapp
during a sled run.
Worked example 2.3.2
CHANGE IN SPEED AND VELOCITY PART 2
A ball is dropped onto a concrete floor and strikes the floor at 5.0 m s−1. It then
rebounds at 5.0 m s−1. The contact with the floor lasts for 25 milliseconds. What
is the average acceleration of the ball during its contact with the floor?
Thinking
Note the values you will need in order
to find the average acceleration (initial
velocity, final velocity and time).
Convert milliseconds into seconds by
dividing by 1000. (Note that ∆v was
calculated for this situation in the
previous Worked example.)
Substitute the values into the average
acceleration equation.
Working
u = −5 m s−1
−u = 5 m s−1
v = 5 m s−1
∆v = 10 m s−1 up
Δt = 25 ms
= 0.025 s
aav =
=
change in velocity
time taken
∆v
∆t
=
10
0.025
= 400ms –2
Acceleration is a vector, so you must
include a direction in your answer.
66
MODULE 1 | KINEMATICS
aav = 400 m s−2 up
Worked example: Try yourself 2.3.2
CHANGE IN SPEED AND VELOCITY PART 2
A ball is dropped onto a concrete floor and strikes the floor at 9.0 m s−1. It then
rebounds at 7.0 m s−1. The contact time with the floor is 35 ms. What is the
average acceleration of the ball during its contact with the floor?
2.3 Review
SUMMARY
• Change in speed is a scalar calculation:
• Acceleration is usually measured in metres per
second per second (m s−2).
Δv = final speed – initial speed = v − u
• Acceleration is a vector. The average acceleration
of a body, aav , is defined as the rate of change of
velocity:
aav = ∆∆vt = v∆−tu
• Change in velocity is a vector calculation:
∆v = final velocity – initial velocity = v − u
KEY QUESTIONS
1
A radio-controlled car is travelling east at 10 km h−1. It
runs over some sand and slows down to 3 km h−1 east.
What is its change in speed?
2
A lump of Blu Tack falling vertically hits the ground
at 5.0 m s−1 without rebounding. What is its change in
velocity during the collision?
3
A ping pong ball hits the floor vertically at 5.0 m s−1
and rebounds directly upwards at 3.0 m s−1. What is its
change in velocity during the bounce?
4
While playing soccer, Ashley is running north at
7.5 m s−1. He slides along the ground and stops in 1.5 s.
What is his average acceleration as he slides to a stop?
5
Olivia launches a model rocket vertically and it
reaches a speed of 150 km h−1 after 3.5 s. What is the
magnitude of its average acceleration in km h−1 s−1?
6
A squash ball travelling east at 25 m s−1 strikes the
front wall of the court and rebounds at 15 m s−1 west.
The contact time between the wall and the ball is
0.050 s. Use vector diagrams, where appropriate, to
help you with your calculations.
a What is the change in speed of the ball?
b What is the change in velocity of the ball?
c What is the magnitude of the average acceleration
of the ball during its contact with the wall?
7
A greyhound starts from rest and accelerates
uniformly. Its velocity after 1.2 s is 8.0 m s−1 south.
a What is the change in speed of the greyhound?
b What is the change in velocity of the greyhound?
c What is the acceleration of the greyhound?
CHAPTER 2 | MOTION IN A STRAIGHT LINE
67
2.4 Graphing position, velocity and
acceleration over time
Sometimes the motion of an object travelling in a straight line is complicated. The
object could travel forwards or backwards, speed up or slow down, stop completely,
or stop and then start again. This information can be presented in a table, or in
graphical form.
Information in a table is not as easy to interpret as information presented graphically.
The main advantage of a graph is that it allows the motion to be visualised clearly. This
section examines position–time, velocity–time and acceleration–time graphs.
POSITION–TIME (x–t) GRAPHS
Position
0
10
20
30
40
50 m
FIGURE 2.4.1 Anneka about to start her swim
from the end of a 50 m pool.
A position–time graph indicates the position x of an object at any time t, for motion
that occurs over an extended time interval.The graph can also show other information.
Consider Anneka, who is attempting to break her personal best for the 50 m
freestyle, as shown in Figure 2.4.1. Her position–time data are shown in Table 2.4.1.
The pool is 50 metres long, and the edge where she dives in is treated as the origin.
TABLE 2.4.1
Anneka’s positions and times during her swim.
Time (s)
0
5
10
15
20
25
30
35
40
45
50
55
60
Position (m)
0
10
20
30
40
50
50
50
45
40
35
30
25
Table 2.4.1 reveals several features of Anneka’s swim. For the first 25 s she swam
at a constant rate. Every 5 s she travelled 10 m in a positive direction, so her velocity
was +2 m s−1. She reached the end of the pool after 25 s. Then from 25 s to 35 s
her position did not change. Finally, from 35 s to 60 s, she swam back towards the
starting point, in a negative direction. On this return lap she maintained a more
leisurely rate of 5 m every 5 s, so her velocity was –1 m s−1.
This data is plotted on the position–time graph in Figure 2.4.2.
50
45
Position (m)
40
35
30
25
20
15
10
5
gradient = velocity
Position (m)
0
Δx
FIGURE 2.4.2
Δt
Time (s)
FIGURE 2.4.3
graph.
68
The gradient of a position–time
MODULE 1 | KINEMATICS
0
5 10 15 20 25 30 35 40 45 50 55 60
Time (s)
The position–time graph for Anneka’s swim.
Anneka’s displacement s can be determined by comparing her initial and final
positions. For example, her displacement between 20 s and 60 s is:
s = final position – initial position
= 25 – 40
= −15 m
By further examining the graph above, it can be seen that during the first 25 s,
Anneka has a displacement of +50 m. This means her average velocity is +2 m s−1,
i.e. 2 m s−1 to the right, during this time. Her velocity can also be obtained by finding
the gradient of this section of the graph (Figure 2.4.3).
SKILLBUILDER
Interpreting the
slope of a linear
graph
The gradient of an x–t graph is the velocity.
A positive velocity indicates that the motion is in a positive direction. A negative
velocity indicates that the motion is in a negative direction.
To confirm that the gradient of a position–time graph is a measure of velocity,
you can use dimensional analysis:
Gradient of x −t graph =
rise
run
Scientists often represent
a relationship between two
variables as a graph. For directly
proportional relationships, the
variables are connected by a
straight line, where the slope (or
gradient) of the line represents
the constant of proportionality
between the two variables.
= ∆∆xt
The units of this gradient is metres per second (m s−1), so gradient is a measure
of velocity. Note that the rise in the graph is the change in position, which is the
definition of displacement; that is, Δx = s .
Non-uniform velocity
For motion with uniform (constant) velocity, the position–time graph will be a
straight line, but if the velocity is non-uniform the graph will be not be straight. If
the position–time graph is curved, as in Figure 2.4.5, the instantaneous velocity (the
velocity at a particular point, such as A) is the gradient of the tangent to the line at
that time. The average velocity between two points (such as B and C) is the gradient
of the chord between the points at the start and end of that period.
The slope or gradient of the
line is defined as the ratio of
change between two points in the
vertical axis (Δy), divided by the
change between two points in the
horizontal axis (Δx). In other words,
it measures the rate at which one
variable (the dependent variable)
changes with respect to the other
(the independent variable).
Displacement
C
B
The graph below has two straight
lines with different slopes. The
steeper slope (blue line) indicates
that the rate of change is higher.
This means the change is
happening more quickly. On the
other hand, the flatter slope (red
line) indicates that the rate of
change is lower. This means the
change is happening more slowly.
A
Time
FIGURE 2.4.5 The instantaneous velocity at point A is the gradient of the tangent at that point. The
average velocity between points B and C is the gradient of the chord between these points on the
graph.
Worked example 2.4.1
y
8
ANALYSING A POSITION–TIME GRAPH
The motion of a cyclist is represented by the position–time graph below. A, B, C, D,
E and F represent points along the cyclist’s journey.
x(m)
D
300
7
6
5
E
4
250
3
200
2
150
1
100
B
50
A
0
20
1
C
40
N
60
80
F
100 t(s)
FIGURE 2.4.4
2
3
x
4
Two lines with different slopes.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
69
a What is the velocity of the cyclist between A and B?
Thinking
Working
Determine the change in position
(displacement) of the cyclist between
A and B using:
s = final position – initial position
At A, x = 0 m.
At B, x = 100 m.
s = 100 − 0
= +100 m or 100 m forwards
(that is, away from the starting point)
Determine the time taken to travel
from A to B.
Δt = 20 − 0
Calculate the gradient of the graph
between A and B using:
gradient of x−t graph = rise
= ∆∆xt
run
Remember that Δx = s.
Gradient = 100
20
State the velocity, using:
Since the gradient is 5, the velocity is
+5 m s−1 or 5 m s−1 forwards.
gradient of x–t graph = velocity
= 20 s
=5
Velocity is a vector so direction must
be given.
b Describe the motion of the cyclist between B and C.
Thinking
Working
Interpret the shape of the graph
between B and C.
The graph is flat between B and C,
indicating that the cyclist’s position
is not changing for this time. So the
cyclist is not moving, and their the
velocity is 0 m s−1.
You can confirm the result by
calculating the gradient of the graph
between B and C using:
Gradient =
gradient of x−t graph = rise
=
run
Remember that Δx = s.
0
20
=0
∆x
∆t
Worked example: Try yourself 2.4.1
ANALYSING A POSITION–TIME GRAPH
Use the graph shown in Worked example 2.4.1 to answer the following questions.
a What is the velocity of the cyclist between E and F?
WS
1.3
70
MODULE 1 | KINEMATICS
b Describe the motion of the cyclist between D and E.
VELOCITY–TIME (v–t) GRAPHS
Analysing motion
Finding displacement
A velocity–time graph can also be used to find the displacement of the object under
consideration.
It is easier to see why the displacement is given by the area under the v–t graph
when velocity is constant. For example, the graph in Figure 2.4.8 shows that in
the first 4 s of motion, Aliyah moved with a constant velocity of +3 m s−1. Note that
the area under the graph for this period of time is a rectangle. Her displacement,
s , during this time can be determined by rearranging the formula for velocity:
v = ∆st
s = v ×t
‒
Velocity (m s–1)
A graph of velocity v against time t shows how the velocity of an object changes with
time. This type of graph is useful for analysing the motion of an object moving in a
complex manner.
Consider the example of Aliyah, who ran backwards and forwards along an aisle
in a store (Figure 2.4.6). Her velocity–time graph shows that she was moving with
a positive velocity, i.e. in a positive direction, for the first 6 s. Between the 6 s mark
and the 7 s mark she was stationary, then she ran in the opposite direction (because
her velocity was negative), for the final 3 s.
The graph shows Aliyah’s velocity at each instant in time. She moved in a
positive direction with a constant speed of 3 m s−1 for the first 4 s. Between 4 s and
6 s she continued moving in a positive direction but slowed down. At 6 s she came
to a stop for 1 s. During the final 3 s she accelerated in the negative direction for 1 s,
then travelled at a constant velocity of −1 m s−1 for 1 s. She then slowed down and
came to a stop at 10 s. Remember that whenever the graph is below the time axis the
velocity is negative, which indicates travel in the opposite direction. So Aliyah was
travelling in the opposite direction for the last 3 s.
+
3
2
1
0
–1
–2
Time (s)
1 2
3 4 5
6 7 8 9 10
FIGURE 2.4.6 Diagram and v–t graph for a girl
running along an aisle.
Displacement s is given by the
area under a velocity–time graph;
that is, the area between the
graph and the time axis.
Note that an area below the
time axis indicates a negative
displacement; that is, motion in a
negative direction.
area = displacement
= height × base
v (m s –1)
= are under v –t graph
Velocity (m s–1)
3
2
1
t (s)
area
FIGURE 2.4.7 The area under a v–t graph
equals the displacement.
= +12 m
= displacement
0
–1
FIGURE 2.4.8
Time (s)
1
2
3
4
5
6
1_
area = × 2 × 3
2

∴ s = +3 m
7
8
9
10
11
Area values as shown in Aliyah’s v–t graph.
From Figure 2.4.8, the area under the graph for the first 4 seconds gives Aliyah’s
displacement during this time, i.e. +12 m. The displacement from 4 s to 6 s is
represented by the area of the darker blue triangle and is equal to +3 m. The total
displacement during the first 6 s is +12 m + 3 m = +15 m.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
71
Worked example 2.4.2
ANALYSING A VELOCITY–TIME GRAPH
The motion of a radio-controlled car, initially travelling east in a straight line across
a driveway, is represented by the graph below.
Velocity (m s–1)
8
6
4
2
0
–2
1
2
3
4
5
6
7
8
9 Time (s)
–4
a What is the displacement of the car during the first 4 seconds?
Thinking
Working
Displacement is the area
under the graph.
Use displacement = b × h for
squares and rectangles.
Use displacement = 12 (b × h)
for triangles.
Velocity (m s–1)
So calculate the area under
the graph for the time period
for which you want to find the
displacement.
8
6
4
Area
2
= +16 m
0
1
–2
2
3
4
5
6
The area from 0 to 4 s is a triangle, so:
=
1
2
×4×8
= +16m
displacement = 16 m east
b What is the average velocity of the car for the first 4 seconds?
Thinking
Identify the equation and variables,
and apply the sign convention.
Working
v = ∆st
s = +16m
∆t = 4 s
Substitute values into the equation:
v = ∆st
v=
=
s
∆t
+16
4
= +4ms−1
Velocity is a vector quantity, so a
direction is needed.
72
MODULE 1 | KINEMATICS
8
–4
area = 12 (b × h)
Displacement is a vector
quantity, so a direction is
needed.
7
v av = 4 m s –1 east
9 Time (s)
Worked example: Try yourself 2.4.2
ANALYSING A VELOCITY–TIME GRAPH
Use the graph from Worked example 2.4.2 to answer the following questions.
a What is the displacement of the car from 4 to 6 seconds?
b What is the average velocity of the car from 4 to 6 seconds?
FINDING ACCELERATION FROM A v–t GRAPH
The acceleration of an object can also be found from a velocity–time graph.
Consider the motion of Aliyah in the 2 s interval between 4 s and 6 s on the graph
in Figure 2.4.9. She is moving in a positive direction, but slowing down from 3 m s−1
until she comes rest.
gradient = acceleration
v (m s –1)
rise
run
t (s)
FIGURE 2.4.9
Gradient as displayed in a v–t graph.
The gradient of a velocity–time graph gives the average acceleration of the
object over the time interval.
Velocity (m s–1)
3
2
1
gradient =
area
= +12 m
= displacement
0
Time (s)
1
2
–1
FIGURE 2.4.10
–3
__
= –1.5 m s–2
2
= acceleration
3
4
5
6
1_
area = × 2 × 3
2
∴ s = +3 m
7
8
9
10
11
Acceleration as displayed in Aliyah’s v–t graph.
Her acceleration is:
v −u
0−3
a = ∆∆vt = ∆t = 2 = −1.5 ms−2
Because acceleration is change in velocity divided by change in time, it is given
by the gradient of the v–t graph. As you can see from Figure 2.4.10, the gradient of
the line between 4 s and 6 s is −1.5 m s−2.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
73
Worked example 2.4.3
FINDING ACCELERATION USING A v–t GRAPH
Consider the motion of the same radio-controlled car initially travelling east in a
straight line across a driveway as shown by the graph below.
Velocity (m s–1)
8
6
4
2
0
–2
1
2
3
4
5
6
7
9 Time (s)
8
–4
What is the acceleration of the car during the first 4 s?
Working
Acceleration is the gradient
of a v–t graph. Calculate the
gradient using:
gradient =
rise
run
8
Velocity (m s–1)
Thinking
6
gradient = –2 m s–2
4
2
0
–2
1
2
3
4
5
6
7
8
9 Time (s)
–4
Gradient from 0 − 4 =
=
rise
run
−8
4
= −2 m s−2
Acceleration is a vector
quantity, so a direction is
needed.
Acceleration = −2 m s−2 east (or 2 m s−2 west)
Note: In this case, the car
is moving in the easterly
direction and slowing down.
Worked example: Try yourself 2.4.3
FINDING ACCELERATION USING A v–t GRAPH
Use the graph shown in Worked example 2.4.3. What is the acceleration of the
car during the period from 4 to 6 seconds?
74
MODULE 1 | KINEMATICS
DISTANCE TRAVELLED
A velocity–time graph can be used to calculate the distance travelled. The process of
determining distance requires you to calculate the area under the v–t graph, similar
to when calculating displacement. However, because distance travelled by an object
always increases as the object moves, regardless of direction, you must add up all the
areas between the graph and the time axis, regardless of whether the area is above
or below the axis.
For example, Figure 2.4.11 shows the velocity–time graph of the radio-controlled
car in Worked example 2.4.3. The area above the time axis, which corresponds to
motion in the positive direction, is +16 m, while the area below the axis, which
corresponds to negative motion, consists of two parts with areas −4 m and −12 m.
To calculate the total displacement, add up the displacements:
total displacement = 16 + (−4) + (−12)
= 16 − 16
= 0m
To calculate the total distance, add up the magnitudes of the areas, ignoring
whether they are positive or negative:
total distance = 16 + 4 + 12
= 32 m
8
Velocity (m s–1)
6
4
2
0
–2
–4
FIGURE 2.4.11
time graph.
gradient = –2 m s–2
area = +16 m
1
2
area = – 4 m
3
4
5
6
gradient = –2 m s–2
7
8
9 Time (s)
area = –12 m
Both distance and displacement can be calculated from the areas under the velocity–
Non-uniform acceleration
For motion with uniform (constant) acceleration, the velocity–time graph is a
straight line. For acceleration that is not uniform, the velocity–time graph is curved.
If the velocity–time graph is curved, the instantaneous acceleration is the gradient of
the tangent to the line at the point of interest; the average acceleration is the gradient
of the chord between two points. The displacement can still be calculated by finding
the area under the graph, but you will need to make some estimations.
ACCELERATION–TIME (a–t) GRAPHS
An acceleration–time graph shows the acceleration of an object as a function of time.
The area under an acceleration–time graph is found by multiplying the acceleration
Δt during which the acceleration occurs. The area gives a change in
a by the time
velocity ∆v value:
area = a × ∆t = ∆v
CHAPTER 2 | MOTION IN A STRAIGHT LINE
75
In order to establish the actual velocity of an object, its initial velocity must
be known. Figure 2.4.12 shows Aliyah’s velocity versus time (v–t) graph and
acceleration versus time (a–t) graph.
‒
+
(a)
Velocity (m s–1)
3
2
1
0
–1
Time (s)
1
2
3
4
5
6
7
8
9
10
–2
(b)
1.5
area = +1 m s–1
Acceleration (m s–2)
1
0.5
0
−0.5
−1
−1.5
−2
PA
1.1
WS
1.4
76
MODULE 1 | KINEMATICS
WS
1.5
FIGURE 2.4.12
1
2
3
4
5
6
7
8
9
10
Time (s)
area = −1 m s–1
area = −3 m s–1
(a) Aliyah’s v–t graph. (b) Aliyah’s a–t graph.
Between 4 s and 6 s the area shows ∆v = –3 m s−1. This indicates that Aliyah has
slowed down by 3 m s−1 during this time. Her v−t graph confirms this fact. Her initial
speed is 3 m s−1, so she must be stationary (v = 0 m s−1) after 6 s. This calculation
could not be made without knowing Aliyah’s initial velocity.
2.4 Review
SUMMARY
• A position–time graph can be used to determine the
location of an object at any given time. Additional
information can also be derived from the graph:
• The gradient of a velocity–time graph is the
acceleration of the object.
• The area under a velocity–time graph is the
displacement of the object.
-- Displacement is given by the change in position
of an object.
• The area under an acceleration–time graph is the
change in velocity of the object.
-- The velocity of an object is given by the
gradient of the position–time graph.
-- If the position–time graph is curved, the
gradient of the tangent at a point gives the
instantaneous velocity.
KEY QUESTIONS
1
Which of the following does the gradient of a position–
time graph represent?
A displacement
B acceleration
C time
D velocity
The following information relates to questions 2–6. The
graph represents the straight-line motion of a radiocontrolled toy car.
6
During its 10 s motion, what was the car’s:
a distance travelled?
b displacement?
7
The velocity–time graphs for a bus and a bicycle
travelling along the same straight stretch of road are
shown below. The bus is initially at rest and starts
moving as the bicycle passes it at time t = 0 s.
12
Velocity (m s–1)
8
Position (m)
6
4
2
0
–2
–4
8
2
3
4
5
6
7
8
9
10
Describe the motion of the car in terms of its position.
3
What was the position of the toy car after:
a 2 s?
b 4 s?
c 6 s?
d 10 s?
4
When did the car return to its starting point?
5
What was the velocity of the toy car:
a during the first 2 s?
b at 3 s?
c from 4 s to 8 s?
d at 8 s?
e from 8 s to 9 s?
0
2
4
6
8 10 12 14 16 18 20 22 24
Time (s)
a What is the magnitude of the initial acceleration of
the bus?
b At what time does the bus overtake the bicycle?
c How far has the bicycle travelled before the bus
catches it?
d What is the average velocity of the bus during the
first 8 s?
Time (s)
2
bicycle
4
0
1
bus
8
a Draw an acceleration–time graph for the bus in
question 7.
b Use your acceleration–time graph to find the change
in velocity of the bus over the first 8 s.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
77
2.5 Equations of motion
DS
1.1
A graph is an excellent way of representing motion because it provides a great deal
of information that is easy to interpret. However, a graph can take a long time to
draw, and sometimes values have to be estimated rather than calculated precisely.
In Section 2.4, graphs of motion were used to evaluate quantities such as
displacement, velocity and acceleration. Here you will use a more powerful and
precise method of solving problems involving constant or uniform acceleration in
a straight line. This method involves the use of a series of equations that can be
derived from the basic definitions developed earlier.
DERIVING THE EQUATIONS
Consider an object moving in a straight line with an initial velocity u and a uniform
acceleration a for a time interval Δt. Because u, v and a are vectors and the motion
is limited to one dimension, the sign and direction convention of right as positive
and left as negative can be used.
After a period of time Δt, the object has changed its velocity from an initial
velocity u and is now travelling with a final velocity v. Its acceleration is given by:
a = ∆∆vt = v∆−tu
If the initial time is 0 s and the final time is t s, then Δt = t. The above equation
can then be rearranged as:
v = u + at Equation 1
The average velocity of the object is:
= ∆st
average velocity vav = displacement
time taken
When acceleration is uniform, the average velocity vav is the average of the initial
and final velocities:
vav = 12 (u + v )
This relationship is shown graphically in Figure 2.5.1.
Velocity (m s–1)
10
average
speed
5
Time (s)
FIGURE 2.5.1
78
MODULE 1 | KINEMATICS
Uniform acceleration displayed by a v–t graph.
4
So:
s
t
= 12 (u + v )
v
This gives:
v–u
Equation 2
A graph describing motion with constant acceleration is shown in Figure 2.5.2. For
constant acceleration, the velocity is increasing by the same amount in each time
interval, so the gradient of the v–t graph is constant.
The displacement s of the body is given by the area under the velocity–time
graph. The area under the velocity–time graph, as shown in Figure 2.5.2, is given by
the combined area of the rectangle and the triangle:
Area = s = ut + 12 × (v − u ) × t
Because a = v −t u
then: v = u + at, and this can be substituted for v − u :
s = ut + 12 × at × t
1 2
s = ut + 2 at Velocity
1 s = 2 ( u + v ) t 1
2
area = _ (v – u)t
u
u
area = ut
t
0
Time
t
FIGURE 2.5.2 The area under a v–t graph
divided into a rectangle and a triangle.
Equation 3
Making u the subject of Equation 1 gives:
u = v − at
You might like to derive another equation yourself by substituting this into
Equation 2. You will get:
1 2 s = v t − 2 at
Equation 4
Rewriting Equation 1 with t as the subject gives:
t = v −a u
Now, if this is substituted into Equation 2:
= u +2 v × v a− u
2 2
= v 2−au
Finally, transposing this gives:
v 2 = u 2 + 2 as Equation 5
CHAPTER 2 | MOTION IN A STRAIGHT LINE
79
SOLVING PROBLEMS USING EQUATIONS
v = u + at
1 s = 2 ( u + v )t
1 2
s = ut + 2 at
1 2
s = v t − 2 at
v 2 = u 2 + 2as
where s is the displacement
in metres
u is the initial velocity in m s−1
v is the final velocity in m s−1
a is the acceleration in m s−2
t is the time taken in seconds.
When you are solving problems using these equations, it is important to think about
the problem and try to visualise what is happening. Follow the steps below.
Step 1 Draw a simple diagram of the situation.
Step 2Write down the information that has been given in the question. You might
like to use the word ‘suvat’ as a memory trick to help you remember to
list the variables in the order s , u, v , a and t. Use a sign convention to
assign positive and negative values to indicate directions. Convert units if
necessary (e.g. from km h−1 to m s−1).
Step 3 Select the equation that matches your data. It should include three values
that you know, and the one value that you want to solve.
Step 4 Use the appropriate number of significant figures in your answer.
Step 5 Include units with the answer, and specify a direction if the quantity is a
vector.
Worked example 2.5.1
USING THE EQUATIONS OF MOTION
A snowboarder in a race is travelling 10 m s−1 north as he crosses the finishing line.
He then decelerates uniformly, coming to a stop over a distance of 20 m.
a What is his acceleration as he comes to a stop?
Thinking
Working
Write down the known quantities as
well as the quantity you are finding.
Take all the information that you can
from the question:
Apply the sign convention that north is
positive and south is negative.
•constant acceleration, so use
equations for uniform acceleration
•‘coming to a stop’ means that the
final velocity is zero.
s = +20 m
u = +10 m s−1
v = 0 m s−1
a =?
Identify the correct equation to use.
Substitute known values into the
equation and solve for a.
Include units with the answer.
v 2 = u2 + 2as
v 2 = u2 + 2as
02 = 102 + 2 × a × 20
0 = 100 + 40 a
−100 = 40 a
a = −100
40
= −2.5 m s−2
Use the sign convention to state the
answer with its direction.
80
MODULE 1 | KINEMATICS
a = 2.5 m s−2 south
b How long does he take to come to a stop?
Thinking
Working
Write down the known quantities as
well as the quantity you are finding.
Take all the information that you can
from the question:
Apply the sign convention that north is
positive and south is negative.
•constant acceleration, so use
equations for uniform acceleration
•‘coming to a stop’ means that the
final velocity is zero.
s = +20 m
u = +10 m s−1
v = 0 m s−1
a = −2.5 m s−2
t=?
Identify the correct equation to use.
Since you now know four values, any
equation involving t will work.
Substitute known values into the
equation and solve for t.
Include units with the answer.
v = u + at
v = u + at
0 = 10 + (−2.5) × t
−10 = −2.5t
10
t = −−2.5
= 4.0 s
c What is the average velocity of the snowboarder as he comes to a stop?
Thinking
Working
Write down the known quantities
as well as the quantity that you are
finding.
Take all the information that you can
from the question:
Apply the sign convention that north is
positive and south is negative.
Identify the correct equation to use.
Substitute known values into the
equation and solve for v av.
Include units with the answer.
Use the sign convention to state the
answer with its direction.
•constant acceleration, so we only
need to find the average of the final
and initial speeds.
u = +10 m s−1
v = 0 m s−1
v av = ?
v av = 12 (u + v )
v av = 12 (u + v )
= 12 (0 + 10)
= 5.0 m s−1
v av = 5.0 m s−1 north
CHAPTER 2 | MOTION IN A STRAIGHT LINE
81
Worked example: Try yourself 2.5.1
USING THE EQUATIONS OF MOTION
A snowboarder in a race is travelling 15 m s−1 east as she crosses the finishing line.
She then decelerates uniformly until coming to a stop over a distance of 30 m.
PA
1.2
PA
1.3
PA
1.4
PA
1.2
PA
1.3
PA
1.4
a What is her acceleration as she comes to a stop?
b How long does she take to come to a stop?
c What is the average velocity of the snowboarder as she comes to a stop?
2.5 Review
SUMMARY
• The following equations can be used for situations
where there is a uniform acceleration, where:
-- s = displacement (m)
-- u = initial velocity (m s−1)
-- v = final velocity (m s−1)
-- a = acceleration (m s−2)
-- t = time (s).
• v = u + at
• s = 12 (u + v )t
2
• s = ut + 12 at
2
• s = vt − 12 at
2 2
• v = u + 2as
• v av = ∆st = u 2+ v
• A sign and direction convention for motion in one
dimension needs to be used with these equations.
KEY QUESTIONS
1
A cyclist has a uniform acceleration as he rolls down a
hill. His initial speed is 5 m s−1, he travels a distance of
30 m and his final speed is 18 m s−1. Which equation
should be used to determine his acceleration?
A v = u + at
B s = 12 (u + v )t
C s = ut + 12 at 2
D s = vt − 12 at 2
E v 2 = u2 + 2as
2
82
A new-model race car travels with a uniform
acceleration on a racetrack. It starts from rest and
covers 400 m in 16 s.
a What is the average acceleration during this time?
b What is the final speed of the car in m s−1?
c What is the car’s final speed in km h−1?
MODULE 1 | KINEMATICS
3
A hybrid car starts from rest and accelerates uniformly
in a positive direction for 8.0 s. It reaches a final speed
of 16 m s−1.
a What is the acceleration of the car?
b What is the average velocity of the car?
c What is the distance travelled by the car?
4
During its launch phase, a rocket accelerates uniformly
from rest to 160 m s−1 upwards in 4.0 s, then travels with
a constant speed of 160 m s−1 for the next 5.0 s.
a What is the initial acceleration of the rocket?
b How far (in km) does the rocket travel in this
9.0 s period?
c What is the final speed of the rocket in km h−1?
d What is the average speed of the rocket during the
first 4.0 s?
e What is the average speed of the rocket during the
9.0 s motion?
5
While overtaking another cyclist, Ben increases his
velocity uniformly from 4.2 m s−1 to 6.7 m s−1 east over
a time interval of 0.50 s.
a What is Ben’s average acceleration during this time?
b How far does Ben travel while overtaking?
c What is Ben’s average speed during this time?
6
A stone is dropped vertically into a lake. Which one of
the following statements best describes the motion of
the stone at the instant it enters the water?
A Its velocity and acceleration are both downwards.
B It has an upwards velocity and a downwards
acceleration.
C Its velocity and acceleration are both upwards.
D It has a downwards velocity and an upwards
acceleration.
7
A diver plunges headfirst into a diving pool while
travelling at 4.4 m s−1 vertically downwards. The diver
enters the water and stops after a distance of 4.0 m.
Consider the diver to be a single point located at her
centre of mass, and assume her acceleration through
the water is uniform.
a What is the average acceleration of the diver as she
travels through the water?
b How long does the diver take to come to a stop?
c What is the velocity of the diver after she has dived
through 2.0 m of water?
8
A car is travelling east along a straight road at
75 km h−1. In an attempt to avoid an accident, the
driver has to brake suddenly and stop the car.
a What is the car’s initial speed in m s−1?
b If the reaction time of the motorist is 0.25 s, what
distance does the car travel while the driver is
reacting to apply the brakes?
c Once the brakes are applied, the car has an
acceleration of −6.0 m s−2. How far does the car
travel while pulling up?
d What total distance does the car travel from the
time the driver first reacts to the danger to when
the car comes to a stop?
CHAPTER 2 | MOTION IN A STRAIGHT LINE
83
2.6 Vertical motion
Until about 500 years ago it was widely believed that heavier objects fall faster than
lighter objects. This was the theory of Aristotle, and it lasted for 2000 years until the
end of the Middle Ages. In the 17th century the Italian scientist Galileo conducted
experiments that showed that the mass of the object did not affect the rate at which
it fell, as long as air resistance was not a factor.
It is now known that falling objects speed up because of gravity. Many people
still think that heavier objects fall faster than light objects. This is not the case, but
the confusion arises because of the effects of air resistance. This section examines
the motion of falling objects.
ANALYSING VERTICAL MOTION
FIGURE 2.6.1 The time elapsed between each
image of the free-falling apple is the same, but
the distance it travels increases between each
image, which shows the apple is accelerating.
Without air resistance, this rate of acceleration is
the same for all objects.
Some falling objects such as feathers and balloons are affected by air resistance to
a large extent. This is why these objects do not speed up much as they fall. However,
if air resistance can be ignored, all free-falling bodies near the Earth’s surface will
move with the same downwards acceleration.
The image in Figure 2.6.1 shows an apple accelerating as it falls, since the
distance travelled by the apple between each photograph increases. In a vacuum,
this rate of acceleration would be the same for a feather, a bowling ball, or any other
object. The mass of the object does not matter.
At the Earth’s surface, the acceleration due to gravity, g, is 9.8 m s−2 downwards,
and does not depend on whether the body is moving upwards or downwards.
As an example, a coin that is dropped from rest will be moving at 9.8 m s−1 after
1 s, 19.6 m s−1 after 2 s, and so on. Each second its velocity increases by 9.8 m s−1
downwards. This is illustrated in Figure 2.6.2.
However, if the coin was thrown straight up at 19.6 m s−1, then after 1 s its speed
would be 9.8 m s−1, and after 2 s it would be stationary. In other words, each second
it would slow down by 9.8 m s−1. The motion of a coin thrown vertically upwards is
shown in Figure 2.6.3.
v=0
9.8 m s–1
v=0
t = 0: at rest
9.8 m s–1
1s
19.6 m s–1
84
MODULE 1 | KINEMATICS
A falling coin.
1s
19.6 m s–1
t=0
2s
FIGURE 2.6.2
2s
FIGURE 2.6.3
A coin thrown vertically upwards.
So regardless of whether the coin is falling or rising, its speed changes at the
same rate. The speed of the falling coin increases by 9.8 m s−1 each second and the
speed of the rising coin decreases by 9.8 m s−1 each second. The acceleration of the
coin due to gravity is 9.8 m s−2 downwards in both cases.
Because the acceleration of a free-falling body is constant, it is appropriate to use
the equations for uniform acceleration. It is necessary to specify whether up or down
is positive when doing these problems. You can simply follow the mathematical
convention of regarding up as positive, which would mean the acceleration due to
gravity is −9.8 m s−2.
PHYSICSFILE N
Acceleration due to gravity
The acceleration due to gravity on the
surface of the Earth, g, varies slightly
−2
from 9.8 m s according to the location.
The reasons for this will be studied in
Year 12 Physics. On the Moon gravity is
much weaker than on Earth, and falling
objects accelerate at only 1.6 m s−2. Other
planets and bodies in the solar system
have different gravity values, depending
on their mass and size. The value of
g at various locations is provided in
Table 2.6.1.
TABLE 2.6.1 Acceleration due to gravity at
different locations on Earth, and on other
bodies in the solar system.
Location
PHYSICSFILE ICT
Galileo’s experiment on the
Moon
In 1971, astronaut David Scott went
to great lengths to show that Galileo’s
prediction was correct. On the Apollo 15
Moon mission he took a hammer and
a feather on the voyage. He stepped
onto the lunar surface, held the feather
and hammer at the same height and
dropped them together. As Galileo
had predicted 400 years earlier, in
the absence of any air resistance the
two objects fell side by side as they
accelerated towards the Moon’s surface.
Acceleration due to
gravity (m s−2)
Sydney
9.797
South Pole
9.832
Equator
9.780
Moon
1.600
Mars
3.600
Jupiter
24.600
Pluto
0.670
FIGURE 2.6.4 Astronaut David Scott holding
a feather and a hammer on the Moon.
Worked example 2.6.1
VERTICAL MOTION
A construction worker accidentally knocks a brick
from a building, and it falls
vertically a distance of 50 m to the ground. Use g = −9.8 m s−2 and ignore air
resistance when answering these questions.
a How long does the brick take to fall halfway, to 25 m?
Thinking
Working
Write down the values of the quantities
that are known and what you are finding.
The brick starts at rest so u = 0.
s = −25 m
u = 0 m s−1
a = −9.8 m s−2
Apply the sign convention that up is
positive and down is negative.
Select the equation for uniform
acceleration that best fits the data
you have.
Substitute known values into the equation
and solve for t.
Think about whether the value seems
reasonable.
t=?
12
s = ut + 2 at
−25 = 0 × t + 12 × −9.8 × t 2
= −4.9t2
t=
−25
−4.9
= 2.3 s
CHAPTER 2 | MOTION IN A STRAIGHT LINE
85
b How long does the brick take to fall all the way to the ground?
Thinking
Write down the values of the quantities
that are known and what you are finding.
Apply the sign convention that up is
positive and down is negative.
Identify the correct equation of uniform
acceleration to use.
Working
s = −50 m
u = 0 m s−1
a = −9.8 m s−1
t=?
12
s = ut + 2 at
Substitute known values into the equation
and solve for t.
−50 = 0 × t + 12 × −9.8 × t 2
Think about whether the value seems
reasonable.
−50 = −4.9t2
Notice that the brick takes 2.3 s to travel
the first 25 m and only 0.9 s to travel the
final 25 m. This is because it is accelerating.
t=
−50
−4.9
= 3.2 s
c What is the velocity of the brick as it hits the ground?
Thinking
Write down the values of the quantities
that are known and what you are finding.
Apply the sign convention that up is
positive and down is negative
Identify the equation for uniform
acceleration that best fits the data you
have.
Substitute known values into the equation
and solve for v .
Think about whether the value seems
reasonable.
Use the sign and direction convention to
describe the direction of the final velocity.
Working
s = −50 m
u = 0 m s−1
v =?
a = −9.8 m s−1
t = 3.2 s
v = u + at
v = 0 + (−9.8) × 3.2
= −31 m s−1
v = −31 m s−1 or 31 m s−1 downwards
Worked example: Try yourself 2.6.1
VERTICAL MOTION
A construction worker accidentally knocks a hammer from a building, and it falls
vertically a distance of 60 m to the ground. Use g = −9.8 m s−2 and ignore air
resistance when answering these questions.
a How long does the hammer take to fall halfway, to 30 m?
b How long does the hammer take to fall all the way to the ground?
c What is the velocity of the hammer as it hits the ground?
When an object is thrown vertically up into the air, it will eventually reach a
point where it stops momentarily before returning downwards. So the velocity of
the object decreases as the object rises, becomes zero at the maximum height, and
then increases again in the opposite direction as the object falls. Throughout this
motion the acceleration due to gravity is −9.8 m s−2. Knowing that the velocity of
an object thrown in the air is zero at the top of its flight allows you to calculate the
maximum height reached.
86
MODULE 1 | KINEMATICS
Worked example 2.6.2
MAXIMUM HEIGHT PROBLEMS
On winning a tennis match the victorious player, Michael, smashed the ball
vertically into the air at 27.5 m s−1. In this example, air resistance can be ignored
and the acceleration due to gravity is –9.80 m s−2.
a Determine the maximum height reached by the ball above its starting
position.
Thinking
Write down the values of the quantities
that are known and what you are
finding. At the maximum height the
velocity is zero.
Apply the sign convention that up is
positive and down is negative.
Select an appropriate formula.
Substitute known values
into the
equation and solve for s.
Working
u = 27.5 m s−1
v =0
a = −9.80 m s−2
s =?
v 2 = u2 + 2a s
0 = (27.5)2 + 2 × (−9.8) × s
s = −756.25
−19.6
s
∴ = +38.6 m
The ball reaches a height of 38.6 m
above its starting position.
b Calculate the time that the ball takes to return to its starting position.
Thinking
To work out the time for which the ball
is in the air, it is often necessary to first
calculate the time that it takes to reach
its maximum height.
Write down the values of the quantities
that are known and what you are
finding.
Select an appropriate formula.
Substitute known values into the
equation and solve for t.
Working
u = 27.5 m s−1
v = 0 m s−1
a = −9.80 m s−2
s = 38.6
t=?
v = u + at
0 = 27.5 + (−9.8 × t)
9.8t = 27.5
∴ t = 2.81 s
The ball takes 2.81 s to reach its
maximum height. It will therefore take
2.81 s to fall from this height back to
its starting point, so the whole trip will
last for 5.62 s.
Worked example: Try yourself 2.6.2
MAXIMUM HEIGHT PROBLEMS
On winning a cricket match, a fielder throws a cricket ball vertically into the air at
15.0 m s−1. In this example, air resistance can be ignored and the acceleration due
to gravity is –9.80 m s−2.
a Determine the maximum height reached by the ball above its starting position.
b Calculate the time that the ball takes to return to its starting position.
WS
1.6
CHAPTER 2 | MOTION IN A STRAIGHT LINE
87
2.6 Review
SUMMARY
• The equations for uniform acceleration can be
used to solve vertical motion problems. It is
necessary to specify whether up or down is
positive.
• If air resistance can be ignored, all bodies falling
freely near the Earth will move with the same
constant acceleration.
• The acceleration due to gravity is 9.8 m s−2 in the
direction towards the centre of the Earth. It is
represented by the symbol g.
KEY QUESTIONS
For these questions, ignore the effects of air resistance and
assume that the acceleration due to gravity is 9.8 m s−2.
1
2
3
88
Angus inadvertently drops an egg while baking a cake,
and the egg falls vertically towards the ground. Which
one of the following statements correctly describes
how the egg falls?
A The egg’s acceleration increases.
B The egg’s acceleration is constant.
C The egg’s velocity is constant.
D The egg’s acceleration decreases.
Max is an Olympic trampolinist who is practising
some routines. Which one or more of the following
statements correctly describes Max’s motion when he
is at the highest point of a bounce? Assume that his
motion is vertical.
A He has zero velocity.
B His acceleration is zero.
C His acceleration is upwards.
D His acceleration is downwards.
A window cleaner working on a skyscraper accidentally
drops her mobile phone. The phone falls vertically
towards the ground with an acceleration of 9.8 m s−2.
a Determine the speed of the phone after 3.0 s.
b How fast is the phone moving after it has fallen
30 m?
c What is the average velocity of the phone during a
fall of 30 m?
MODULE 1 | KINEMATICS
4
A rubber ball is bounced so that it travels straight up
into the air, reaching its highest point after 1.5 s.
a What is the initial velocity of the ball just as it leaves
the ground?
b What is the maximum height reached by the ball?
5
A book is knocked off a bench and falls vertically to
the floor. If the book takes 0.40 s to fall to the floor,
calculate the following values.
a What is the book’s speed as it lands?
b What is the height from which the book fell?
c How far did the book fall during the first 0.20 s?
d How far did the book fall during the final 0.20 s?
6
While celebrating her birthday, Mishti pops a party
popper. The lid travels vertically into the air. Being a
keen physics student, Mishti notices that the lid takes
4.0 s to return to its starting position.
a How long did the lid take to reach its maximum
height?
b How fast was the lid travelling initially?
c What was the maximum height reached by the lid?
d What was the velocity of the lid as it returned to its
starting point?
Chapter review
KEY TERMS
acceleration
air resistance
centre of mass
collinear
dimension
dimensional analysis
direction convention
displacement
distance travelled
free-fall
magnitude
newton
position
rectilinear
resultant
scalar
speed
unit
vector
vector diagram
vector notation
velocity
KEY QUESTIONS
1
2
Select the scalar quantities in the list below. (There
may be more than one answer.)
A force
B time
C acceleration
D mass
Select the vector quantities in the list below.
(There may be more than one answer.)
A displacement
B distance
C volume
D velocity
10 A car travels at 95 km h−1 along a freeway. What is its
speed in m s−1?
11 A cyclist travels at 15 m s−1 during a sprint finish. What
is this speed in km h−1?
The following information relates to questions 12 and 13.
An athlete in training for a marathon runs 15 km north
along a straight road before realising that she has dropped
her drink bottle. She turns around and runs back 5 km
to find her bottle, then resumes running in the original
direction. After running for 2.0 hours, the athlete reaches
20 km from her starting position and stops.
12 Calculate the average speed of the athlete in km h−1.
3
A basketballer applies a force with his hand to bounce
the ball. Describe how a vector can be drawn to
represent this situation.
13 Calculate her average velocity in:
a km h−1
b m s−1.
4
Vector arrow A is drawn twice the length of vector
arrow B. What does this mean?
5
A car travels 15 m s−1 north and another travels
20 m s−1 south. Why is a sign convention often used to
describe vectors like these?
14 A ping pong ball is falling vertically at 6.0 m s−1 as it
hits the floor. It rebounds at 4.0 m s−1 up. What is its
change in speed during the bounce?
6
If the vector 20 N forwards is written as −20 N, how
would you write a vector representing 80 N backwards?
7
Add the following force vectors using a number
line: 3 N left, 2 N right, 6 N right. Then also draw and
describe the resultant force vector.
8
Determine the resultant vector of the following motion:
45.0 m forwards, then 70.5 m backwards, then 34.5 m
forwards, then 30.0 m backwards.
9
Determine the change in velocity of a bird that
changes from flying 3 m s−1 to the right to flying 3 m s−1
to the left.
15 A car is moving in a positive direction. It approaches
a red light and slows down. Which of the following
statements correctly describes its acceleration and
velocity as it slows down?
A The car has positive acceleration and negative
velocity.
B The car has negative acceleration and positive
velocity.
C Both the velocity and acceleration of the car are
positive.
D Both the velocity and acceleration of the car are
negative.
16 A skier is travelling along a horizontal ski run at a
speed of 15 m s−1. After falling over, the skier takes 2.5 s
to come to rest. Calculate the average acceleration of
the skier.
CHAPTER 2 | MOTION IN A STRAIGHT LINE
89
CHAPTER REVIEW CONTINUED
17 The following graph shows the position of a motorcycle
along a straight stretch of road as a function of time.
The motorcycle starts 200 m north of an intersection.
19 The following velocity–time graph is for an Olympic
road cyclist as he travels, initially north, along a
straight section of track.
14
500
12
Velocity (m s–1)
400
Position (m)
300
200
100
10
20
30
40
50
60 Time (s)
–200
a During what time interval is the motorcycle
travelling north?
b During what time interval is the motorcycle
travelling south?
c During what time intervals is the motorcycle
stationary?
d At what time is the motorcycle passing back
through the intersection?
18 For each of the situations listed below, indicate which
of the velocity–time graphs best represents the motion
involved.
A
B
v
v
t
t
D
v
t
C
t
E
t
a A car comes to a stop at a red light.
b A swimmer is travelling at a constant speed.
c A motorbike starts from rest with uniform
acceleration.
90
6
4
0
–100
v
8
2
0
v
10
MODULE 1 | KINEMATICS
0
1
2
3
4
5
6
7
Time (s)
8
9
10 11 12
a Calculate the displacement of the cyclist during his
journey.
b Calculate the magnitude, to three significant figures,
of the average velocity of the cyclist during this
11.0 s interval.
c Calculate the acceleration of the cyclist at t = 1 s.
d Calculate the acceleration of the cyclist at t = 10 s.
e Which one or more of the following statements
correctly describes the motion of the cyclist?
A He is always travelling north.
B He travels south during the final 2 s.
C He is stationary at t = 8 s.
D He returns to the starting point after 11 s.
20 A car starts from rest and has a constant acceleration
of 3.5 m s−2 for 4.5 s. What is its final speed?
21 A jet-ski starts from rest and accelerates uniformly. If it
travels 2.0 m in its first second of motion, calculate:
a its acceleration
b its speed at the end of the first second
c the distance the jet-ski travels in its second second
of motion.
22 A skater is travelling along a horizontal skate rink at a
speed of 10 m s−1. After falling over, she takes 10 m to
come to rest. Calculate, to two significant figures, the
answers to the following questions about the skater’s
movement.
a What is her average acceleration?
b How long does it take her to come to a stop?
23 The following graph shows the position of Candice as
she dances across a stage.
B
Position (m)
A
C
D
12
8
4
0
0
5
10
15
Time (s)
20
25
a What is Candice’s starting position?
b In which of the sections A–D is Candice at rest?
c In which of the sections A–D is Candice moving in a
positive direction, and what is her velocity?
d In which of the sections A–D is Candice moving with
a negative velocity and what is the magnitude of
this velocity?
e Calculate Candice’s average speed during the 25 s
motion.
24 Anna is cycling at a constant speed of 12 m s−1 when
she passes a stationary bus. The bus starts moving
just as Anna passes, and it accelerates uniformly at
1.5 m s−2.
a When does the bus reach the same speed as Anna?
b How long does the bus take to catch Anna?
c What distance has Anna travelled before the bus
catches up?
For the following questions, the acceleration due to gravity
is 9.8 m s−2 down and air resistance is considered to be
negligible.
25 Two physics students conduct the following
experiment from a very high bridge. Thao drops a
1.5 kg sphere from a height of 60.0 m, while at exactly
the same time Benjamin throws a 100 g cube with an
initial downwards velocity of 10.0 m s−1 from a point
10.0 m above Thao.
a How long does it take the sphere to reach the
ground?
b How long does it take the cube to reach the ground?
26 At the start of an AFL football match, the umpire
bounces the ball so that it travels vertically upwards
and reaches a height of 15.0 m.
a How long does the ball take to reach this
maximum height?
b One of the ruckmen is able to leap and reach to a
height of 4.0 m with his hand. How long after the
bounce should this ruckman try to make contact
with the ball?
27 A billiard ball rolls from rest down a smooth ramp that
is 8.0 m long. The acceleration of the ball is constant
at 2.0 m s−2.
0
4.0 m
8.0 m
a What is the speed of the ball when it is halfway
down the ramp?
b What is the final speed of the ball?
c How long does the ball take to roll the first 4.0 m?
d How long does the ball take to travel the final 4.0 m?
28 Four metal bolts are tied to a piece of rope. The rope is
dropped and the metal bolts hitting the ground create
a steady rhythm, making a sound at 0.25 second
intervals. Calculate the distances between each of the
metal bolts.
29 After completing the activity on page 63, reflect on
the following inquiry question: How is the motion
of an object moving in a straight line described and
predicted?
CHAPTER 2 | MOTION IN A STRAIGHT LINE
91
CHAPTER
CHAPTER
Motion on a plane
Using the concept of vectors, this chapter will analyse the motion of objects moving
in two dimensions. You will learn how to break a vector down into its components
and then analyse the relative motion of objects such as aeroplanes flying through
strong winds or boats travelling across rivers with fast currents.
Content
INQUIRY QUESTION
How is the motion of an object that changes its direction of
movement on a plane described?
By the end of this chapter you will be able to:
• analyse vectors in one and two dimensions to:
-- resolve a vector into two perpendicular components
-- add two perpendicular vector components to obtain a single vector
(ACSPH061) N
• represent the distance and displacement of objects moving on a horizontal plane
using:
-- vector addition
-- resolution of components of vectors (ACSPH060) ICT N
• describe and analyse algebraically, graphically and with vector diagrams, the ways
in which the motion of objects changes, including: ICT
-- velocity
-- displacement (ACSPH060, ACSPH061) N
• describe and analyse, using vector analysis, the relative positions and motions of
one object relative to another object on a plane (ACSPH061)
• analyse the relative motion of objects in two dimensions in a variety of situations,
for example:
-- a boat on a flowing river relative to the bank
-- two moving cars
-- an aeroplane in a crosswind relative to the ground
(ACSPH060, ACSPH132). ICT N
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
PHYSICS INQUIRY
N
CCT
Wind-assisted art
How is the motion of an
object that changes its
direction of movement on
a plane described?
COLLECT THIS…
•
marbles or ball bearings, all the
same size and mass
•
paint (3 or 4 different colours)
•
3 or 4 plates
•
roll of paper
•
fan with different speed settings
3.1 Vectors in two dimensions
When motion is in one dimension, it is relatively simple to understand direction.
However, some motions require a description in a two-dimensional plane. You might
need to describe the motion of somebody walking up hill, a movement which includes
walking horizontally (forwards) and vertically (up). In this section you will learn how
to apply the rules of addition and subtraction to vectors in two dimensions.
VECTORS IN TWO DIMENSIONS
Just as in one-dimensional vector analysis, it is helpful to use certain direction
conventions to describe the two-dimensional planes. These planes could be:
• horizontal, which is commonly defined using north, south, east and west
• vertical, which can be defined in a number of ways including forwards, backwards,
up, down, left and right.
The description of the direction of these vectors is more complicated. Therefore,
a more detailed convention is needed for identifying the direction of a vector. There
are a variety of conventions, but they all describe a direction as an angle from a
known reference point.
DO THIS…
Horizontal plane
1 Set up the fan so that air will
blow across a table.
For a horizontal, two-dimensional plane, the two common methods for describing
the direction of a vector are:
• full circle (or true) bearing — A ‘full circle bearing’ describes north as zero
degrees true. This is written as 0°T. In this convention, all directions are given as
a clockwise angle from north. As an example, 95°T is 95° clockwise from north.
• quadrant bearing — An alternative method is to provide a ‘quadrant bearing’,
where all angles are referenced from either north or south and are between 0°
and 90° towards east or west. In this method, 30°T becomes N30°E, which can
be read as ‘from north 30° towards the east’.
Using these two conventions, north-west (NW) would be 315°T using a full
circle bearing, or N45°W using a quadrant bearing. Figure 3.1.1 demonstrates
these two methods.
2 Tape the piece of paper down
to the table to prevent it from
flapping.
3 Place a tablespoon of paint
on each plate. Roll the marble
in the paint so that it is all
covered.
4 With the fan on the lowest
setting, roll the marble along
the paper.
5 Repeat with the same colour,
moving down the paper each
time. Try to roll all marbles at
the same speed.
north
north-west or
315°T or
N 45°W
45°
6 Repeat with different fan
settings, changing the colour of
the paint to match the different
fan settings.
north-east or
45°T or
N 45°E
45°
west
east
315°
RECORD THIS…
Describe what shape the paint
created.
Present a diagram of your
investigation.
REFLECT ON THIS…
How is the motion of an object that
changes its direction of movement
on a plane described?
Where else have you seen
movement that created similar
paths?
94
MODULE 1 | KINEMATICS
south
FIGURE 3.1.1 Two horizontal vector directions, viewed from above, using full circle bearings and
quadrant bearings.
Vertical plane
For a vertical, two-dimensional plane the directions are referenced to vertical (upwards
and downwards) or horizontal (left and right) and are between 0° and 90° clockwise or
anticlockwise. For example, a vector direction can be described as ‘60° clockwise from
the left direction’. The same vector direction could be described as ‘30° anticlockwise
from the upwards direction’. The opposite direction to this vector would be ‘60°
clockwise from the right direction’. This example is illustrated in Figure 3.1.2.
Worked example 3.1.1
30° anticlockwise from the upwards direction
or
DESCRIBING TWO-DIMENSIONAL VECTORS
60° clockwise from the left direction
up
Describe the direction of the vector using an appropriate method.
up
30°
70°
left
right
left
60°
60°
right
30°
down
down
60° clockwise from the right direction
or
Thinking
Working
Choose the appropriate points to
reference the direction of the vector.
In this case using the vertical reference
makes more sense, as the angle is
given from the vertical.
The vector can be referenced to the
vertical.
Determine the angle between the
reference direction and the vector.
In this example there is 70° from the
vertical to the vector.
Determine the direction of the vector
from the reference direction.
From vertically up, the vector is
clockwise.
Describe the vector using the sequence:
angle, clockwise or anticlockwise from
the reference direction.
This vector is 70° clockwise from the
upwards direction.
30° anticlockwise from the downwards direction
FIGURE 3.1.2
Two vectors in the vertical plane.
Worked example: Try yourself 3.1.1
DESCRIBING TWO-DIMENSIONAL VECTORS
Describe the direction of the following vector using an appropriate method.
up
left
50°
right
down
ADDING VECTORS IN TWO DIMENSIONS
Adding vectors in two dimensions means that all of the vectors must be in the same
plane. The vectors can go in any direction within the plane, and can be separated by
any angle. The examples in this section illustrate vectors in the horizontal plane, but
the same strategies apply to adding vectors in the vertical plane.
CHAPTER 3 | MOTION ON A PLANE
95
The horizontal plane is one that is looked down on from above. Examples include
looking at a house plan or map placed on a desk. The direction conventions that
suit this plane best are the north, south, east and west convention, or the forwards,
backwards, left and right convention. These are shown in Figure 3.1.3.
north
forwards
west
east
left
right
south
FIGURE 3.1.3
backwards
The direction conventions for the horizontal plane.
Graphical method of adding vectors
The magnitude and direction of a resultant vector can be determined by measuring
an accurately drawn scaled vector diagram. There are two main ways to do this:
• head to tail method
• parallelogram method.
Head to tail method
To add vectors at right angles to each other using a graphical method, use an
appropriate scale and then draw each vector head to tail. The resultant vector is
the vector that starts at the tail of the first vector and ends at the head of the last
vector. To determine the magnitude and direction of the resultant vector, measure
the length of the resultant vector and compare it to the scale, then measure and
describe the direction appropriately.
In Figure 3.1.4, two vectors, 30.0 m east and 20.0 m south, are added head to
tail. The resultant vector, shown in red, is measured to be about 36 m according
to the scale provided. Using a protractor, the resultant vector is measured to be in
the direction 34° south of east. This represents a direction of S 56° E when using
quadrant bearings.
30.0 m east
N
W
E
S
0
160
20
170
10
0
40
14
40
50
13
0
120
60
110
70
100
80
90
80
100
70
110
50
60
1
20
0
14
30
0
150
30
15
0
13
MODULE 1 | KINEMATICS
10
170
20
160
96
0
180
180
10 m
FIGURE 3.1.4
20.0 m south
R = 36 m
E 34° S
Adding two vectors at right angles, using the graphical method.
If the two vectors are at angles other than 90° to each other, the graphical
method is ideal for finding the resultant vector. In Figure 3.1.5, the vectors 15 N
east and 10 N S45°E are added head to tail. The magnitude of the resultant vector
is measured to be about 23 N. The direction of the resultant vector is measured by a
protractor from east to be 18° towards the south, which should be written as S72°E.
15 N east
N
W
E
10 N S45°E
S
→
R = 23 N
0
10
20
160
160
20
170
10
170
50
13
0
120
60
110
70
100
80
90
80
100
70
110
60
50
0
12
1
40
40
0
40
14
150
30
0
30
15
0
13
FIGURE 3.1.5
0
180
180
5N
E 18° S
Adding two vectors not at right angles, using the graphical method.
Parallelogram method
An alternative method for determining a resultant vector is to construct a
parallelogram of vectors. In this method the two vectors to be added are drawn tail
to tail. Next, a parallel line is drawn for each vector as shown in Figure 3.1.6. In
this figure, the parallel lines have been drawn as dotted lines. The resultant vector is
drawn from the tails of the two vectors to the intersection of the dotted parallel lines.
→
1
→
1
v
→
v2
FIGURE 3.1.6
→
R
v
→
v2
→
R
Parallelogram of vectors method for adding two vectors.
Geometric method of adding vectors
Graphical methods of adding vectors in two dimensions only give approximate
results as they rely on comparing the magnitude of the resultant vector to a scale
and measuring the direction with a protractor. A more accurate method to resolve
vectors is to use Pythagoras’ theorem and trigonometry. These techniques are
referred to as geometric methods. Geometric methods can be used to calculate
the magnitude of the vector and its direction. However, Pythagoras’ theorem and
trigonometry can only be used for finding the resultant vector of two vectors that
are at right angles to each other.
SKILLBUILDER
N
Understanding
Pythagoras’
theorem
For any right-angled triangle,
Pythagoras’ theorem states that
the square of the length of the
hypotenuse is equal to the sum of
the squares of the lengths of the
two shorter sides.
a2 + b2 = c2
A
b
C
c
a
B
CHAPTER 3 | MOTION ON A PLANE
97
In Figure 3.1.7, two vectors, 30.0 m east and 20.0 m south, are added head to tail.
The resultant vector, shown in red, is calculated using Pythagoras’ theorem to be
36.1 m. The resultant vector is calculated to be in the direction S56.3°E. This result
is more accurate than the answer determined earlier in this section.
N
W
30.0 m east
θ
E
S
→
R
R 2 = 30.02 + 20.02
= 900 + 400
20.0
tan θ = 30.0
θ = tan–1 0.6667
R = √1300
SKILLBUILDER
= 36.1 m
N
FIGURE 3.1.7
Understanding
sine, cosine
and tangent
relationships
in right-angled
triangles
cos (θ ) =
tan(θ ) =
opposite
adjacent
Determine the child’s resultant displacement vector.
Refer to Figure 3.1.3 on page 96 for sign and direction conventions if required.
Thinking
Construct a vector diagram
showing the vectors drawn
head to tail. Draw the
resultant vector from the tail of
the first vector to the head of
the last vector.
sin(θ )
cos (θ )
tan(θ )
Acronym
SOH
CAH
TOA
In the triangle shown, if two
variables are known then it is
possible to calculate all of the other
values. For example, if you have the
values of angle B and the length of
the hypotenuse, you can calculate b
using the formula:
=
Working
N
W
E
S
16.0 m north
s→
25.0 m west
Equations
opposite
hypotenuse
Adding two vectors at right angles, using the geometric method.
A child runs 25.0 m west and 16.0 m north.
The acronyms for each of these
rules are:
sin B =
E33.7°S (or S56.3°E)
ADDING VECTORS IN TWO DIMENSIONS USING GEOMETRY
opposite
hypotenuse
adjacent
hypotenuse
= 33.7°
Worked example 3.1.2
You will recall that:
sin(θ ) =
20.0 m south
A
As the two vectors to be added
are at 90° to each other,
apply Pythagoras’ theorem to
calculate the magnitude of the
resultant displacement.
Using trigonometry, calculate
the angle from the west vector
to the resultant vector.
s2 = 25.02 + 16.02
= 625 + 256
s = 881
= 29.7 m
tanθ =
16
25
θ = tan−1 0.64
= 32.6°
Determine the direction of the
vector relative to north or south.
State the magnitude and
direction of the resultant vector.
b
c
θ
90° − 32.6° = 57.4°
The direction is N57.4°W
s = 29.7 m N57.4°W
b = sin B × c
b
Worked example: Try yourself 3.1.2
c
ADDING VECTORS IN TWO DIMENSIONS USING GEOMETRY
Forces of 5.0 N east and 3.0 N north act on a tree.
C
98
a
MODULE 1 | KINEMATICS
B
Determine the resultant force vector acting on the tree.
Refer to Figure 3.1.3 on page 96 for sign and direction conventions if required.
SUBTRACTING VECTORS IN TWO DIMENSIONS
Changing velocity in two dimensions can occur when turning a corner; for
example, walking at 3 m s−1 west, then turning to travel at 3 m s−1 north. Although
the magnitude of the velocity is the same, the direction is different.
A change in velocity in two dimensions can be determined using either the
graphical method or the geometric method similar to vector addition. The initial
velocity must always be reversed before it is added to the final velocity.
The two-dimensional direction conventions were introduced earlier and are
shown here in Figure 3.1.8.
north
west
forwards
east
left
south
FIGURE 3.1.8
A change in a vector can be
caused by a change in magnitude,
a change in direction, or a change
in both magnitude and direction.
right
backwards
The direction conventions for the horizontal plane.
Graphical method of subtracting vectors
To subtract vectors using a graphical method, use a direction convention and a scale
and draw each vector.
Using velocity as an example, the steps to do this are as follows:
• Draw in the final velocity first.
• Draw the opposite of the initial velocity head to tail with the final velocity vector.
• Draw the resultant change in velocity vector, starting at the tail of the final velocity
vector and ending at the head of the opposite of the initial velocity vector.
• Measure the length of the resultant vector and compare it to the scale to
determine the magnitude of the change in velocity.
• Measure an appropriate angle to determine the direction of the resultant vector.
Figure 3.1.9 shows the velocity vectors for travelling 3 m s−1 west and then
turning and travelling 3 m s−1 north. The opposite of the initial velocity is drawn as
3 m s−1 east.
N
W
v2 = 3 m s–1 north
→
E
S
1 m s–1
FIGURE 3.1.9
v1 = 3 m s–1 west
→
–→v1 = 3 m s–1 east
Subtracting two vectors at right angles, using the graphical method.
CHAPTER 3 | MOTION ON A PLANE
99
To determine the change in velocity, the final velocity vector is drawn first.
Then from its head the opposite of the initial velocity is drawn. This is shown in
Figure 3.1.10. The magnitude of the change in velocity (resultant vector) is shown
in red. It is measured to be about 4.3 m s−1 according to the scale provided. Using a
protractor, the resultant vector is measured to be in the direction N45°E.
3 m s–1 east
0 10
20
180 170
160
3 m s north
–1
∆v = 4.3 m s
N 45° E
80
70
0
60 110 10
0
50 0 1 2
13
–1
→
30
15
0 40
14
0
90
Subtracting two vectors at right angles, using the graphical method.
100
80
FIGURE 3.1.10
Geometric method of subtracting vectors
The graphical method of subtracting vectors in two dimensions only gives
approximate results, as it relies on comparing the magnitude of the change in
velocity vector to a scale and measuring its direction with a protractor.
A more accurate method to subtract vectors is to use Pythagoras’ theorem and
trigonometry.
Figure 3.1.11 shows how to calculate the resultant velocity when changing from
25 m s−1 east to 20.0 m s−1 south. The initial velocity of 25.0 m s−1 east and the final
velocity of 20.0 m s−1 south are drawn. Then the opposite of the initial velocity is drawn
as 25.0 m s−1 west. The final velocity vector is drawn first, then from its head the opposite
of the initial velocity is drawn. The resultant velocity vector, shown in red, is calculated
to be 32.0 m s−1. The resultant vector is calculated to be in the direction S51.3°W.
The resultant vector is 32.0 m s−1 S51.3°W.
v1 = 25.0 m s–1 east
→
N
W
v2 = 20.0 m s–1 south
→
E
–v→ 1 = 25.0 m s–1 west
S
θ
∆→v
v2 = 20.0 m s–1 south
→
–→v1 = 25.0 m s–1 west
R2 = 25.02 + 20.02
= 625 + 400
R = √1025
= 51.3°
= 32.0 m s
–1
FIGURE 3.1.11
100
MODULE 1 | KINEMATICS
25.0
20.0
θ = tan–1 1.25
tan θ =
S 51.3° W
Subtracting two vectors at right angles, using the geometric method.
Worked example 3.1.3
SUBTRACTING VECTORS IN TWO DIMENSIONS USING GEOMETRY
Clare approaches a corner on her scooter at 18.7 m s−1 west, and exits the corner
at 16.6 m s−1 north.
Determine Clare’s change in velocity.
Thinking
Draw
the final velocity vector,
v2, and the initial velocity
vector, v1, separately. Then
draw the initial velocity in the
opposite direction.
Working
N
E
W
S
v→ 1 = 18.7 m s–1 west
v2 = 16.6 m s–1 north
→
Construct a vector diagram
drawing v2 first and then from
its head draw the opposite of v1.
The change of velocity vector
is drawn from the tail of the
final velocity to the head of the
opposite of the initial velocity.
–v→ 1 = 18.7 m s–1 east
N
W
E
–v→ 1 = 18.7 m s–1 east
S
v→ 2 = 16.6 m s–1 north
∆→v
θ
As the two vectors to be added
are at 90° to each other,
apply Pythagoras’ theorem to
calculate the magnitude of the
change in velocity.
Calculate the angle from the
north vector to the change in
velocity vector.
State the magnitude and
direction of the change in
velocity.
∆v 2 = 16.62 + 18.72
= 275.26 + 349.69
∆v = 625.25
= 25.0 m s−1
18.7
tanθ = 16.6
θ = tan−1 1.13
= 48.4°
∆v = 25.0 m s−1 N48.4°E
Worked example: Try yourself 3.1.3
SUBTRACTING VECTORS IN TWO DIMENSIONS USING GEOMETRY
A ball hits a wall at 7.0 m s−1 south and rebounds at 6.0 m s−1 east.
Determine the change in velocity of the ball.
CHAPTER 3 | MOTION ON A PLANE
101
PHYSICS IN ACTION
ICT
Surveying
Surveyors use technology to measure, analyse and manage
data about the shape of the land and the exact location of
landmarks and buildings. They take many measurements,
including angles and distances, and use them to calculate
more advanced data such as vectors, bearings, co-ordinates,
elevations, maps etc. Surveyors typically use theodolites
(Figure 3.1.12 and Figure 3.1.13), GPS survey equipment,
laser range finders and satellite images to map the land in
three dimensions.
FIGURE 3.1.12
Surveying the land with a theodolite.
Surveyors are often the first professionals who work
on a building site, to ensure that the boundaries of the
property are correct. They also ensure that the building
is built in the correct location. Surveyors must liaise
closely with architects both before and during a building
project as they provide position and height data for walls
and floors.
FIGURE 3.1.13
Surveying equipment being used on a building site.
3.1 Review
SUMMARY
• Adding and subtracting vectors in two dimensions
can be estimated graphically with a scale and a
protractor.
• Adding and subtracting vectors in two dimensions
can be calculated using Pythagoras’ theorem and
the trigonometric ratios of a right-angled triangle.
• An alternative method of adding vectors in two
dimensions is to construct a parallelogram of vectors.
KEY QUESTIONS
1
A jet plane in level flight makes a turn, changing
its velocity from 345 m s−1 south to 406 m s−1 west.
Calculate the change in the velocity of the jet.
2
Yvette hits a golf ball that strikes a tree and changes
its velocity from 42.0 m s−1 east to 42.0 m s−1 north.
Calculate the change in the velocity of the golf ball.
3
A yacht tacks during a race, changing its velocity from
7.05 m s−1 south to 5.25 m s−1 west. Calculate the
change in the velocity of the yacht.
4
102
Describe the magnitude and direction of the resultant
vector, drawn in red, in the following diagram.
MODULE 1 | KINEMATICS
θ
→
R
20 m south
40 m west
5
Forces of 2000 N north and 6000 N east act on an
object. What is the resultant force?
3.2 Vector components
Section 3.1 explored how vectors can be combined to find a resultant vector. In
physics there are times when it is useful to break one vector up into two vectors that
are at right angles to each other. For example, if a force vector is acting at an angle
up from the horizontal, as shown in Figure 3.2.1, this vector can be considered to
consist of two independent vertical and horizontal components.
The components of a vector can be found using trigonometry.
→
F = 45 N
FIGURE 3.2.1 The pulling force acting on the cart has a component in the horizontal direction and a
component in the vertical direction.
45 N
Vectors at an angle are more easily dealt with if they are broken up into perpendicular
components; that is, two components that are at right angles to each other. These
components, when added together, give the original vector. To find the components
of a vector, a right-angled triangle is constructed with the original vector as the
hypotenuse. This is shown in Figure 3.2.2. The hypotenuse is always the longest
side of a right-angled triangle and is opposite the 90° angle. The other two sides of
the triangle are each shorter than the hypotenuse and form the 90° angle with each
other. These two sides are the perpendicular components of the original vector.
→
FINDING PERPENDICULAR COMPONENTS OF A VECTOR
F V = 45 sin 20°
= 15 N
20°
20°
→
F h = 45 cos 20°
= 42 N
FIGURE 3.2.2 The perpendicular components
(shown in red) of the original vector (shown in
blue). The original vector is the hypotenuse of
the triangle.
Geometric method of finding vector components
The geometric method of finding the perpendicular components of vectors is to
construct a right-angled triangle using the original vector as the hypotenuse. This
was illustrated in Figure 3.2.2. The magnitude and direction of the components are
then determined using trigonometry. A good rule to remember is that no component
of a vector can be larger than the vector itself. In a right-angled triangle, no side is
longer than the hypotenuse. The original vector must be the hypotenuse and its
components must be the other two sides of the triangle.
Figure 3.2.3 shows a force vector of 50.0 N (drawn in black) acting on a box
in a direction 30.0° up from the horizontal to the right. The horizontal and vertical
components of this force must be found in order to complete further calculations.
→
F = 50.0 N
→
F vertical
30.0°
→
F horizontal
FIGURE 3.2.3
Finding the horizontal and vertical components of a force vector.
CHAPTER 3 | MOTION ON A PLANE
103
The horizontal component vector is drawn from the tail of the 50.0 N vector
towards the right, with its head directly below the head of the original 50.0 N vector.
The vertical component vector is drawn from the head of the horizontal component
to the head of the original 50.0 N vector.
Using trigonometry, the horizontal component of the force is calculated to be
43.3 N horizontally to the right. The vertical component is calculated to be 25.0 N
vertically upwards. The calculations are shown below:
PHYSICSFILE
Complex motion
When looking at more complex motion,
such as the projectile motion of a ball
thrown through the air, the vector
components become particularly
important. When you throw a ball, the
initial forward force from your arm
moves the ball across horizontally. At
the same time, the weight force of the
ball moves the ball down vertically. It is
the addition of these two components
which gives the complex shape you see
when you throw a ball.
cosθ =
adjacent
hypotenuse
adj = hyp × cos θ
Fh = 50.0 × cos 30.0°
= 43.3 N horizontal to the right
opposite
hypotenuse
opp = hyp × sin θ
Fv = 50.0 × sin 30.0°
= 25.0 N vertically upwards
Worked example 3.2.1
CALCULATING THE PERPENDICULAR COMPONENTS OF A FORCE
A 235 N force acts on a bike in a direction 17.0° north of west.
Use the direction conventions to determine the perpendicular components of
the force.
Thinking
Draw FW from the tail of the 235 N
force alongthe horizontal direction,
then draw FN from the horizontal
vector to the head of the 235 N force.
FIGURE 3.2.4 Adding together the vectors
for the forward motion of the ball plus the
downward motion of the weight create this
parabola type shape.
sin θ =
Calculate
the west component of the
force FW using
cos θ =
adjacent
hypotenuse
Working
W
N
E
S
→
FN
→
F = 235 N
→
FW
17.0°
adjacent
cos θ = hypotenuse
adj = hyp × cos θ
FW = 235 × cos 17.0°
= 224.7 = 225 N west
Calculate
the north component of the
force FN using
sinθ =
opposite
hypotenuse
sinθ =
opposite
hypotenuse
opp = hyp × sin θ
FN = 235 × sin 17.0°
= 68.7 N north
Worked example: Try yourself 3.2.1
CALCULATING THE PERPENDICULAR COMPONENTS OF A FORCE
A 3540 N force acts on a trolley in a direction 26.5° anticlockwise from the left
direction.
WS
1.2
104
MODULE 1 | KINEMATICS
WS
1.8
Use the direction conventions to determine the perpendicular components of
the force.
3.2 Review
SUMMARY
• Perpendicular component vectors are at right
angles to each other.
• A right-angled triangle vector diagram can be
drawn with the original vector as the hypotenuse
and the perpendicular components drawn from
the tail of the original to the head of the original.
• Any component vectors must be smaller in
magnitude than the original vector.
• The perpendicular components can be
determined using trigonometry.
• A vector can be resolved into two perpendicular
component vectors.
• The hypotenuse of a right-angled triangle is the
longest side of the triangle, and the other two
sides are each smaller than the hypotenuse.
KEY QUESTIONS
1
Rayko applies a force of 462 N on the handle of a
mower in a direction of 35.0° clockwise down from the
right direction.
a What is the downwards force applied?
b What is the rightwards force applied?
2
A force of 25.9 N acts in the direction of S40.0°E. Find
the perpendicular components of the force.
3
A ferry is transporting students to Rottnest Island.
At one point in the journey the ferry travels at
18.3 m s−1 N75.6°W. Calculate its velocity in the
northerly direction and in the westerly direction at
that time.
4
Zehn walks 47.0 m in the direction of S66.3°E across
a hockey field. Calculate the change in Zehn’s position
down the field and across the field during that time.
5
A cargo ship has two tugboats attached to it by ropes.
One tugboat is pulling directly north, while the other is
pulling directly west. The pulling forces of the tugboats
combine to produce a total force of 23 000 N in a
direction of N62.5°W. Calculate the force that each
tugboat applies to the cargo ship.
6
Resolve the following forces into their perpendicular
components around the north–south line. In part d,
use the horizontal and vertical directions.
a 100 N S60°E
b 60 N north
c 300 N 160°T
d 3 × 105 N 30° upwards from the horizontal.
CHAPTER 3 | MOTION ON A PLANE
105
3.3 Relative motion
Now that you know how to algebraically work with vectors in two dimensions, it
is possible to analyse and describe the motion of objects moving on a plane. This
section will use some practical examples to further explore motion in two dimensions.
Relative motion refers to the fact that motion may involve different frames
of reference. A frame of reference is where an observation is being made from.
Sometimes this may be a stationary point; for example, standing on a platform
watching a train pull into a station. But at other times the frame of reference may also
be in motion. Take the example of two competing race cars (Figure 3.3.1). It may be
that you want to calculate their velocities relative to the finish line, which is a stationary
point. Or you may want to find the relative velocity between the two moving cars.
TWO MOVING CARS
FIGURE 3.3.1 Two cars race towards the
finish line. The velocity of the each car can be
calculated relative to the ground (stationary
frame of reference), or relative to each other
(moving frame of reference).
→
BG
ν
= 50 km h–1
→
RG
ν
= 60 km h–1
FIGURE 3.3.2 Two cars approaching each other
from different directions.
To find the relative velocity between two moving objects you need to use the concept
of vector addition.
Choosing the right vector notation
In a one-dimensional example, two cars are approaching each other from different
directions (Figure 3.3.2). The velocities of the cars are both given relative to
the ground, which is a stationary frame of reference. The
blue car travels east at
50 km h–1 ( v BG ) and the red car travels at 60 km h–1 west (v RG ).
The notation for each of these vectors gives you a clue as to what they are describing.
For example,
v BG is for the velocity of the blue car (B) relative to the ground (G).
Similarly, v RG is for the velocity of the red car (R) relative to the ground (G).
By using vector addition you can calculate the resultant vector. The following
formula will calculate the velocity of the blue car relative to the red car:
v BR = v BG + vGR
The order of these vectors is important. If you were to look for the velocity of the
red car relative to the blue car you would use the formula:
v RB = v RG + vGB
When finding the relative velocity of object 1 relative to object 3, using a
stationary frame of reference as object 2, the equation needs to be in the order:
v13 = v12 + v 23
Flipping vectors
There may be times that you will need to flip the order of the vector’s subscripts.
Using
the car example, you are given the velocity of the red car relative to the ground
(v RG ) but the equation for v BR requires the velocity of the ground relative to the red
car (vGR ). To find vGR you can use the vector rule:
v AB = −vBA
So the equation for the velocity of the red car relative to the blue car can now be
written as:
v BR = v BG + vGR
= v BG + (−v RG )
= +50 + (− − 60)
= +50 + (+60)
= +110 km h −1
This process can be used to find the relative velocity in two dimensions.
106
MODULE 1 | KINEMATICS
Worked example 3.3.1
FIND THE RELATIVE VELOCITY BETWEEN TWO CARS IN TWO DIMENSIONS
Two cars are both approaching the same intersection with different velocities.
The blue car is travelling at 50 km h–1 towards the east, and the silver car is
travelling 60 km h–1 north.
Find the velocity of the blue car relative to the silver car.
Thinking
Working
Define your vectors with appropriate
notation. Write out the equation for the
velocity of the blue car relative to the
silver car.
vBG = velocity of the blue car relative to
the ground
v SG = velocity of the silver car relative
to the ground
vBS = velocity of the blue car relative to
the silver car
vBS = vBG + vGS
= vBG + (−vSG )
Construct a vector diagram showing
the vectors drawn head to tail. Draw
the resultant vector from the tail of
the first vector to the head of the last
vector.
As the two vectors to be added are at
90° to each other, apply Pythagoras’
theorem to calculate the magnitude of
the resultant velocity.
N
→
BG
ν
W
= 50 km h–1
ν→BS
E
S
→
–νSG = 60 km h–1
2
vBS
= 502 + 602
= 2500 + 3600
vBS = 6100
= 78 km h–1
Using trigonometry, calculate the angle
from the east vector to the resultant
vector.
tanθ =
60
50
θ = tan−1(1.2)
= 50°
Determine the direction of the vector
relative to north or south.
90 – 50 = 40°
State the magnitude and direction of
the resultant vector.
vBS = 78 km h–1, S40°E
Worked example: Try yourself 3.3.1
FIND THE RELATIVE VELOCITY BETWEEN TWO CARS IN TWO DIMENSIONS
A black car and a yellow car are travelling down the same road towards the
south at 55 km h–1. The yellow car turns off onto a side road towards the west
and travels at 70 km h–1.
Find the velocity of the yellow car as it travels on the side road, relative to the
black car.
CHAPTER 3 | MOTION ON A PLANE
107
BOAT ON A RIVER
It is possible to apply these rules of relative velocity to the motion of a boat on a
river. If a boat is travelling across a river with a strong current (Figure 3.3.3), you
will need to add the forward motion of the boat to the velocity of the river current
to find the resultant vector.
Using the same rules described earlier, the boat’s velocity relative to the ground
can be calculated as:
FIGURE 3.3.3 When calculating the resultant
velocity of a boat you must take into
consideration its forward motion and the velocity
of the current.
vBG = vBW + v WG
where vBG is the velocity of the boat relative to the ground
vBW is the velocity of the boat relative to the water
v WG is the velocity of the water relative to the ground.
Worked example 3.3.2
CALCULATE THE RELATIVE VELOCITY OF A BOAT ON A RIVER
A boat is travelling at 5.0 m s–1 north across a river relative to the water. The
current is flowing downstream at 2.0 m s–1 east.
Determine the velocity of the boat relative to the ground, giving the direction to
the nearest degree.
Thinking
Write out the equation describing the
resultant velocity.
Working
vBG = vBW + v WG
→
WG
ν
Construct a vector diagram showing
the vectors drawn head to tail. Draw
the resultant vector from the tail of
the first vector to the head of the last
vector.
As the two vectors to be added are at
90° to each other, apply Pythagoras’
theorem to calculate the magnitude of
the resultant velocity.
Using trigonometry, calculate the angle
from the north vector to the resultant
vector.
→
BW
ν
= 5.0 m s–1
= 2.0 m s–1
→
BG
ν
2
= 5.02 + 2.02
v BG
= 25 + 4
vBG = 29
= 5.4 m s–1
tanθ =
2.0
5.0
θ = tan−1 0.4
= 21.8 = 22°
Determine the direction of the vector
relative to north or south.
The direction is N22°E
State the magnitude and direction of
the resultant vector.
vBG = 5.4 m s–1, N22°E
Worked example: Try yourself 3.3.2
CALCULATE THE RELATIVE VELOCITY OF A BOAT ON A RIVER
A boat is travelling at 3.8 m s–1 south across a river relative to the water. The
current is flowing upstream at 2.0 m s–1 west.
Determine the velocity of the boat relative to the ground, giving the direction to
the nearest degree.
108
MODULE 1 | KINEMATICS
Aeroplane in a cross wind
A similar situation can be applied to calculating the velocity of an aeroplane with
respect to the ground. As an aeroplane flies it will experience winds blowing
opposite to its direction of motion (head wind), in the same direction to its motion
(tail wind), or at some angle across its direction of motion (cross wind). If you
know both the velocity of the plane relative to the wind and the velocity of the wind
relative to the ground, by using the rules for vector addition the resultant vector of
these two values will describe the velocity of the plane relative to the ground.
vPG = vPW + v WG
where vPG is the velocity of the plane relative to the ground
vPW is the velocity of the plane relative to the wind
v WG is the velocity of the wind relative to the ground
Worked example 3.3.3
FIND THE RESULTANT VELOCITY OF AN AEROPLANE IN A CROSS WIND
A light aircraft is travelling at 300 km h–1 north, with a crosswind blowing at
45.0 km h–1 west.
Determine the velocity of the plane relative to the ground.
Thinking
Working
Write out the equation describing the
resultant velocity.
vPG = vPW + v WG
Construct a vector diagram showing
the vectors drawn head to tail. Draw
the resultant vector from the tail of
the first vector to the head of the last
vector.
As the two vectors to be added are at
90° to each other, apply Pythagoras’
theorem to calculate the magnitude of
the resultant velocity.
Using trigonometry, calculate the angle
from the west vector to the resultant
vector.
→
WG
ν
N
W
Components of flight
= 45 km h–1
→
PG
ν
→
PW
ν
= 300 km h–1
2
= 3002 + 452
v PG
= 90 000 + 2025
vPG = 92 025
= 303 km h–1
tanθ =
WS
1.9
PHYSICSFILE ICT N
E
S
WS
1.7
45
300
θ = tan−1 0.15
= 8.53°
Determine the direction of the vector
relative to north or south.
The direction is N8.53°W
State the magnitude and direction of
the resultant vector.
vPG = 303 km h–1, N8.53°W
The velocity vector that describes the
direction and speed of an aeroplane
can be broken down into multiple
vector components. These are known
as thrust, lift, drag and weight forces.
The thrust is generated by the engines
and gives the plane its forward
motion. The weight force describes the
downwards pull due to gravity. The drag
component slows the plane down as
it pushes through the air. And the lift
component is produced by the wings
and makes the plane rise.
When designing a plane, all of these
components need to be considered.
Worked example: Try yourself 3.3.3
FIND THE RESULTANT VELOCITY OF AN AEROPLANE IN A CROSS WIND
A jet aircraft is travelling at 900 km h–1 east, with a crosswind blowing at
85.0 km h–1 south.
FIGURE 3.3.4 There are multiple vector
components involved in the direction and
speed of an aeroplane.
Determine the velocity of the plane relative to the ground.
CHAPTER 3 | MOTION ON A PLANE
109
3.3 Review
SUMMARY
• In order to calculate the relative velocity of two
moving objects, a third stationary reference frame
must be used.
• The frame of reference describes where an
observation is being made from.
• A frame of reference can be stationary or it can be
in motion.
• When finding the relative velocity between objects
A and B using the stationary reference frame C,
the equation will be:
v AB = v AC + vCB
• Relative motion uses vector addition to find the
resultant vector.
• When describing the velocity of object A relative to
object B, the vector notation looks like v AB.
• When working with vectors remember that:
v AB = −vBA
KEY QUESTIONS
5
A small boat pushes off from the edge of a river at
2.5 m s–1 north. The current is running downstream to
the west at 0.8 m s–1. Calculate the resultant velocity of
the boat.
a A plane is travelling south at 910 km h–1 N into a
headwind of 70.0 km h–1. Calculate the resultant
velocity of the plane relative to the ground.
b The wind changes direction and is now blowing
perpendicular to the plane towards the east.
Calculate the resultant velocity of the plane relative
to the ground.
6
A boat pushes off from a pontoon and travels at
5.0 m s–1 to the east, while the pontoon floats away at
2.0 m s–1 south from its starting position. Calculate the
velocity of the boat relative to the pontoon.
Two Formula 1 cars are racing straight towards the
finish line. The velocity of car 1 relative to the ground
is 315 km h–1, and the velocity of car 2 is 319 km h–1.
Calculate the velocity of car 2 relative to car 1.
7
Ellisa is running across a soccer field at 6 m s–1 north.
She kicks the soccer ball to another player and
continues running with the same velocity. The ball
travels at 20 m s–1 east. Calculate the velocity of Ellisa
relative to the ball.
1
The velocity of a car relative to the ground is given by
vCG and the velocity of a train relative to the ground is
given by v TG. Write out the equation to find the velocity
of the car relative to the train.
2
3
4
A plane is flying north even though it is pointing at
some angle towards the east. The velocity of the plane
relative to the wind is 300 km h–1. The crosswind has
a velocity of 50.0 km h–1. Using the diagram below,
calculate the direction the plane is pointing towards in
order for it to fly due north.
→
PW
ν
= 50 km h–1
N
W
ν
ν→PW = 300 km h–1
110
E
S
→
PG
MODULE 1 | KINEMATICS
Chapter review
KEY TERMS
head wind
relative motion
tail wind
component
cross wind
frame of reference
KEY QUESTIONS
1
When finding the change in velocity between an initial
velocity of 34.0 m s−1 south and a final velocity of
12.5 m s−1 east, which two vectors need to be added
together?
2
Describe the following vector direction.
c 90 km h–1
200 km h–1
θ
up
d
θ
12.1 m s
–1
left
right
70°
5
Jordi walks 4.5 m east before turning north and
walking a further 7.2 m north. Using Pythagoras’
theorem, calculate the magnitude of the resultant
displacement vector.
6
Marta rides her bike with a velocity of 6.5 m s−1 N22°W.
Calculate the vector components of her velocity.
7
What is the magnitude of the resultant vector when
30.0 m south and 40.0 m west are added?
A 7.7 m
B 44.7 m
C 50.0 m
D 2000 m
8
What are the horizontal and vertical components of a
300 N force that is applied along a rope at 60° to the
horizontal and used to drag an object across a yard?
9
An aeroplane flies a distance of 300 km due north,
then changes course and travels 400 km due
east. What is the distance travelled and the final
displacement of the aeroplane?
A distance = 700 km, displacement = 500 km
north east
B distance = 700 km, displacement = 700 km
north east
C distance = 700 km, displacement = 500 km
north 53.1° east
D distance = 700 km, displacement = 500 km
north 36.9° east
down
3
Draw the following vector directions.
a N25°E
b 344°T
4
For each of the following, which trigonometry function
(sin, cos or tan) would be needed to find the angle?
a
12.1 m s
–1
25.5 m s–1
θ
b
θ
220 km h–1
200 km h–1
25.5 m s–1
10 Josie runs up a flight of stairs at a speed of 5.2 m s−1.
The stairs have a rise angle of 30°. Calculate the
vertical and horizontal components of her velocity.
CHAPTER 3 | MOTION ON A PLANE
111
CHAPTER REVIEW CONTINUED
11 Aliyah is walking straight up a steep slope. After
she has walked 8.0 m at an angle of 20°, the slope
increases to 30° and she reaches the top in another
8.0 m. Add together the vertical components of her
displacement for each part of the slope to find the
total height of the slope.
12 Jess is swimming across a river with a current of
0.30 m s−1 to the north. Her velocity relative to the
water is 0.80 m s−1 to the east. Calculate her velocity
relative to the ground.
13 The velocity of a plane relative to the ground is given
by vPG and the velocity of a helicopter relative to the
ground is given by vHG. Write out the equation to find
the velocity of the plane relative to the helicopter.
14 A Cessna light plane is flying north at 300 km h−1
relative to the ground, while the plane itself is angled
8° towards the east. Calculate the velocity of the cross
wind relative to the ground.
15 A yacht is travelling across Parramatta River with a
velocity of 8.0 m s−1 north relative to the water. The
current of the river has a velocity of 2.2 m s−1 west
relative to the ground. Calculate the relative velocity of
the yacht to the ground.
16 An aeroplane is flying at 890 km h−1 north and a tail
wind starts to blow at 40 km h−1.
a What direction is the velocity of the tail wind?
b Calculate the new velocity of the aeroplane relative
to the ground.
17 A rowboat is travelling across a river where the current
is 2.4 m s−1 downstream to the west. The boat’s velocity
relative to the water is 3.0 m s−1 north. Calculate the
relative velocity of the boat to the ground.
18 A train and a car are both approaching a level
crossing. The train is travelling at 100 km h−1 east and
the car is travelling at 60 km h−1 south. Calculate the
velocity of the train relative to the car.
112
MODULE 1 | KINEMATICS
19 A marble is rolled across a table and a fan blows
perpendicular to its motion.
a The velocity of the marble is 0.80 m s−1 relative to
the table, at an angle of 10°T. The fan is to the west
of the marble. Calculate the magnitude of the vector
components of the marble’s velocity.
b Using vector notation, describe what each of
these components represents. For example, what
component describes the velocity of the fan’s wind
relative to the table?
20 a A boat pushes straight off from the edge of a river at
4.0 m s−1. The current runs downstream at 1.5 m s−1.
Draw a vector diagram to describe this situation.
b Calculate the velocity of the boat relative to the
ground.
c The trip takes five minutes for the boat to reach
the opposite bank. Assuming that the velocity is
calculated in part b is the average velocity of the
boat, calculate the displacement.
d Calculate the width of the river.
e The captain of the boat had wanted to reach a
point on the bank directly opposite from where they
pushed off from. Because the current moved them
further downstream, calculate how far from their
destination the boat ended up.
21 a Describe how the motion of a ball being thrown
through the air can be broken down into vector
components.
b What shape describes the motion of the ball?
22 After completing the activity on page 94, reflect on
the following inquiry question: How is the motion of
an object that changes its direction of movement on a
plane described?
MODULE 1 • REVIEW
REVIEW QUESTIONS
MR
1
Kinematics
Multiple choice
1
4
−2
A car accelerates in a straight line at a rate of 5.5 m s
from rest. What distance has the car travelled at the end
of three seconds?
A 8.25 m
B 11 m
C 16.5 m
D 24.75 m
2
A bike accelerates in a straight line at a rate of 2.5 m s−2
from rest. What distance does the bike travel in the third
second of its motion?
A 6.25 m
B 13.75 m
C 19.25 m
D 24.75 m
3
A graph depicting the velocity of a small toy train versus
time is shown below. The train is moving on a straight
section of track, and is initially moving in an easterly
direction.
Velocity of train (m s–1)
4
6
8
10
12
Time (s)
A ball is dropped, falls vertically and strikes the ground
with a velocity of +5 m s−1. It rebounds, and leaves the
ground with a velocity of −3 m s−1. What is the change in
velocity that the ball experiences?
A −8 m s−1
B +8 m s−1
C −2 m s−1
D +2 m s−1
6
An aeroplane flies a distance of 300 km due north, then
changes course and travels 400 km due east.
What is the distance travelled and the final displacement
of the aeroplane?
A distance = 700 km, displacement = 500 km north-east
B distance = 700 km, displacement = 700 km north-east
C distance = 700 km, displacement = 500 km N53.1°E
D distance = 700 km, displacement = 500 km N36.9°E
7
A car that is initially at rest begins to roll down a steep
road that makes an angle of 11° with the horizontal.
Assuming a constant acceleration of 2 m s−1, what is the
speed of the cr after it has travelled 100 metres?
A 19 m s−1
B 20 km h−1
C 72 km h−1
D 72 m s−1
8
Which equation can be used to calculate the velocity of
a boat relative to a submarine? (Use the subscripts B for
boat, S for submarine and G for ground.)
A vBS = vBG + v SG
B vBS = vSG + vBS
C vBS = vSG + ( −vBG )
D vBS = vBG + ( −v SG )
–0.1
–0.2
a What distance does the train travel in the first
6 seconds of its motion?
A 0m
B 0.4 m
C 0.8 m
D 1.2 m
b What is the displacement of the train after the first
11 seconds of its motion?
A 0m
B 0.4 m east
C 0.8 m east
D 1.2 m east
A ball dropped from rest from a height h hits the
ground with a speed v. The ball is then released from a
height of 2h. With what speed would the ball now strike
the ground?
A 12 v
5
+0.1
2
WS
1.11
B 2v
C 2v
D 4v
+0.2
0
WS
1.10
REVIEW QUESTIONS
113
MODULE 1 • REVIEW
9
The diagram is an idealised velocity−time graph for the
motion of an Olympic sprinter.
10
9
Velocity (m s–1)
8
7
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15
Time (s)
a What distance was this race?
A 15 m
B 10 m
C 100 m
D 120 m
b What is the average speed of the sprinter for the total
time she is moving?
A 7.0 m s−1
B 8.0 m s−1
C 9.0 m s−1
D 10.0 m s−1
10 The diagram below gives the position–time graph of the
motion of a boy on a bicycle. The boy initially travels in a
northerly direction.
A
90
BC
D
E
F
G
80
Position (m)
70
60
40
30
20
10
0
10 20 30 40 50 60 70 80
Time (s)
During which one or more of the sections, A to G, is the
boy:
a travelling north?
b speeding up?
114
12 A sailboat changes course during a race. It was sailing
east at 6.9 m s−1 but is now sailing north at 7.2 m s−1.
What was the change in velocity of the boat?
A 10 m s−1 N44°E
B 10 m s−1 N46°E
C 10 m s−1 N46°W
D 10 m s−1 N44°W
13 A car is travelling with a velocity of 100 km h−1 S44°W.
What are the vector components of the velocity?
A 70 km h−1 west and 30 km h−1 south
B 70 km h−1 west and 72 km h−1 south
C 72 km h−1 west and 70 km h−1 south
D 30 km h−1 west and 70 km h−1 south
14 Zeke is riding a scooter at a velocity of 6 m s−1 N12°E.
What are the vector components of his velocity?
A 1.2 m s−1 east and 5.9 m s−1 north
B 5.9 m s−1 east and 1.2 m s−1 north
C 4.0 m s−1 east and 2.0 m s−1 north
D 2.0 m s−1 east and 4.0 m s−1 north
15 Jocelyn is swimming across a river in a current of
1.0 m s−1 east. Her velocity relative to the water is
1.3 m s−1 north. What is her velocity relative to the
ground?
A 1.8 m s−1, N52°E
B 1.6 m s−1, N52°E
C 1.8 m s−1, N38°E
D 1.6 m s−1, N38°E
16 Mia is driving alongside a train track at 60 km h−1 east
and there is a train driving in the opposite direction at
150 km h−1 west. What is Mia’s velocity relative to the
train?
A 90 km h−1 east
B 90 km h−1 west
C 210 km h−1 east
D 210 km h−1 west
50
0
11 An apple falls vertically from a tree and hits the top of a
fence at 6.3 m s−1. It bounces off the fence horizontally
with a velocity of 2.1 m s−1. What is the magnitude of the
change in velocity of the apple?
A 3.0 m s−1
B 5.9 m s−1
C 6.6 m s−1
D 8.9 m s−1
MODULE 1 | KINEMATICS
17 A plane is travelling at 600 km h−1 south when a head
wind starts to blow at 38 km h−1. What is the distance
travelled by the plane at the resultant velocity in the next
two hours?
A 562 km
B 1276 km
C 1200 km
D 1124 km
19 Two bikes are approaching an intersection. Briar is riding
at 5.5 m s−1 east and Mei is travelling at 5.1 m s−1 north.
What is the magnitude of the velocity of Briar relative to
Mei?
A 7.5 m s−1
B 10.6 m s−1
C 0.4 m s−1
D 4.2 m s−1
24 The following position versus time graph depicts the
motion of a cyclist travelling east along a straight road
from points A to D.
500
400
350
300
Short answer
21 An Olympic archery competitor tests a bow by firing an
arrow of mass 25 g vertically into the air. The arrow leaves
the bow with an initial vertical velocity of 100 m s−1. The
acceleration due to gravity may be taken as g = −9.8 m s−2
and the effects of air resistance can be ignored.
a At what time will the arrow reach its maximum
height?
b What is the maximum vertical distance that this arrow
reaches?
c What is the acceleration of the arrow when it reaches
its maximum height?
22 A hiker travels 8 km in a northerly direction from his
campsite and then travels a further 7 km in a northeasterly direction.
a What is his final displacement?
b If the journey takes a total of 7 hours, calculate his
the average speed in m s−1.
23 Helen and Ajuna conduct the following experiment from a
skyscraper. Helen drops a platinum sphere from a height
of 122 m. At exactly the same time Ajuna throws a lead
sphere with an initial vertical velocity of −10.0 m s−1 from a
height of 122 m. Assume g = −9.80 m s−2.
a How long does it take the platinum sphere to strike
the ground?
b Calculate the time taken by the lead sphere to strike
the ground.
c Determine the average velocity of each sphere over
their 122 m fall.
C
250
200
150
B
100
−1
20 A light aircraft is flying at 102 km h south and a tail
wind starts to blow at 30 km h−1. What is the new velocity
of the plane relative to the ground?
A 72 km h−1 south
B 132 km h−1 north
C 132 km h−1 south
D 72 km h−1 north
D
450
Position (m)
18 Rachel and Taylor went for a hike in the Blue Mountains.
Using GPS on their phones, they found that after walking
for 2 hours they had travelled a distance of 6 km. What
was their average speed?
A 0.8 m s−1
B 5.0 km h−1
C 3.0 m s−1
D 1.8 km h−1
50
0
A
0
10
20
30
Time (s)
40
50
a
b
c
d
e
Describe the motion of the cyclist in terms of speed.
What was the velocity of the cyclist for the first 30 s?
What was the velocity of the cyclist for the final 10 s?
Calculate the average velocity between points B and C.
Calculate the average acceleration between points B
and C.
f Calculate the average speed between points A and D.
25 A small car is found to slow down from 90 km h−1 to
60 km h−1 in 12 s when the engine is switched off and
the car is allowed to coast on level ground. The car has a
mass of 830 kg.
a What is the car’s deceleration (in m s−2) during the
12 s interval?
b Determine the distance that the car travels during the
12 s interval.
26 A cyclist is riding downhill at 15 m s−1. When they reach
the end of the bike path they slow to a stop over 20 m.
a How long does it take them to come to a complete
stop?
b What is their acceleration over this period?
27 A car with good brakes but smooth tyres has a maximum
retardation of 4.0 m s−2 on a wet road. The driver has a
reaction time of 0.50 s. The car is travelling at 72 km h−1
when the driver sees a danger and reacts by braking.
a How far does the car travel during the reaction time?
b Assuming maximum retardation, calculate the
braking time.
c Determine the total distance travelled by the car from
the time the driver realises the danger to the time the
car finally stops.
REVIEW QUESTIONS
115
MODULE 1 • REVIEW
28 The following displacement−time graph shows the
motion of a person walking.
Displacement (m)
3
2
1
0
–1
2
4
6
8
10
12
14
Time (s)
–2
–3
a Describe the motion of the person using this graph.
b Create a velocity–time graph.
c Create an acceleration–time graph.
29 The current world record time for the 100 m sprint
is 9.58 s.
a What is the average speed over this distance?
b If that speed could be maintained over longer
distances, how long would it take to run 1 km?
30 Gina dives off the 10 m high diving board into a pool.
a Calculate how long it takes her to reach the water.
b What is her final velocity as she enters the water?
Extended response
31 During a physics experiment a student sets a multiflash
timer at a frequency of 10 Hz. A marble is then rolled
across a horizontal table. The diagram shows the
position of the marble for the first four flashes: A, B, C
and D.
Assume that when flash A occurred t = 0, at which time
the marble was at rest.
A
B
C
D
1.0 cm
3.0 cm
5.0 cm
a Determine the average speed of the marble for these
distance intervals:
i A to B
ii B to C
iii C to D
b Determine the instantaneous speeds of the marble
for these times:
i t = 0.05 seconds
ii t = 0.15 seconds
iii t = 0.25 seconds
c Describe the motion of the marble.
116
MODULE 1 | KINEMATICS
32 An Australian yacht and a New Zealand yacht are neckand-neck in the Sydney to Hobart yacht race.
a At one point the Australian yacht is travelling at
exactly 26 km h−1 south and the New Zealand yacht is
sailing at exactly 22 km h−1 east, relatively to the water.
What is the speed of the Australian yacht relative to
the New Zealand yacht?
b Assume the yachts both started travelling in
these directions from the same point. What is the
magnitude of the displacement between the yachts
after 60 seconds travelling at these velocities?
c The velocities given are both relative to the water.
If the current of the water is travelling at exactly
1.5 m s−1 south, calculate the velocity of:
i the New Zealand yacht relative to the land
ii the Australian yacht relative to the land.
d Using the resultant velocities calculated in part c,
and taking the starting point as the displacements
calculated in part b, find the magnitude of the total
displacement between the yachts after they have
travelled for a further 60 seconds.
33 Tessa is canoeing across a river with a current of
2.5 m s−1 to the west.
a Her resultant velocity (relative to the ground) is
3.6 m s−1 north. Calculate the velocity of the canoe
relative to the water.
b The current increases to 3.1 m s−1 west, but Tessa
continues with the same velocity relative to the water.
What is her resultant velocity relative to the ground?
c She continues in the same direction for 1 minute.
Calculate her displacement.
d The river is 219 m across. Would she make it to the
other side?
34 Derive:
a v = u + at
b s = ut + 12 at 2
c v 2 = u2 + 2as
35 Two ships leave port at the same time. Ship 1 heads
south-west with a speed of 50 km h−1. Ship 2 travels at
60 km h−1 N80°E.
a Calculate the velocity of ship 2 relative to ship 1.
b Calculate the velocity of ship 1 relative to ship 2.
c The two ships continue to travel at these velocities.
What is the magnitude of the displacement between
the two ships 1.5 hours after leaving port?
MODULE
Dynamics
The relationship between the motion of objects and the forces that act on them
is often complex. However, Newton’s Laws of Motion can be used to describe the
effect of forces on the motion of single objects and simple systems. This module
develops the key concept that forces are always produced in pairs that act on
different objects and add to zero.
By applying Newton’s laws directly to simple systems, and, where appropriate,
the law of conservation of momentum and law of conservation of mechanical
energy, the effects of forces can be examined. It is also possible to examine the
interactions and relationships that can occur between objects by modelling and
representing these using vectors and equations.
In many situations, within and beyond the discipline of physics, knowing the
rates of change of quantities provides deeper insight into various phenomena.
In this module, the rates of change of displacement, velocity and energy are of
particular significance and students develop an understanding of the usefulness
and limitations of modelling.
Outcomes
By the end of this module you will be able to:
• design and evaluate investigations in order to obtain primary and secondary
data and information PH11-2
• select and process appropriate qualitative and quantitative data and
information using a range of appropriate media PH11-4
• solve scientific problems using primary and secondary data, critical thinking
skills and scientific processes PH11-6
• describe and explain events in terms of Newton’s Laws of Motion, the law of
conservation of momentum and the law of conservation of energy PH11-9
Physics Stage 6 Syllabus © NSW Education Standards Authority for and on
behalf of the Crown in right of the State of NSW, 2017.
CHAPTER
Forces
In the seventeenth century Sir Isaac Newton published three laws that explain why
objects in our universe move as they do. These laws became the foundation of a
branch of physics called mechanics: the science of how and why objects move. They
have become commonly known as Newton’s three laws of motion.
Using Newton’s laws, this chapter will describe the relationship between the forces
acting on an object and its motion.
Content
INQUIRY QUESTION
How are forces produced between objects and what effects do
forces produce?
By the end of this chapter you will be able to:
• using Newton’s Laws of Motion, describe static and dynamic interactions between
two or more objects and the changes that result from:
-- a contact force
-- a force mediated by fields
• explore the concept of net force and equilibrium in one-dimensional and simple
two-dimensional contexts using: (ACSPH050) ICT N
-- algebraic addition
-- vector addition
-- vector addition by resolution into components
• solve problems or make quantitative predictions about resultant and component
forces by applying the following relationships: ICT N
-- F AB = −F BA
-- F x = F cos θ , F y = F sinθ
• conduct a practical investigation to explain and predict the motion of objects on
inclined planes (ACSPH098) CCT ICT
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
4.1 Newton’s first law
PHYSICS INQUIRY
CCT N
Balanced forces
How are forces produced between objects and what effects do forces produce?
COLLECT THIS…
•
3 spring balances
•
steel ring
•
piece of paper
•
large piece of cardboard
•
thumb tack
•
protractor
DO THIS…
1 Draw x and y axes on the centre of the paper. Use the thumb tack to hold the paper to a large piece of cardboard and to
mark the centre.
2 Place the ring over the tack with the spring balances hooked on the ring.
3 Arrange the 3 spring balances so they are equally spaced.
4 With two other people, apply a pulling force to the ring so that it does not move. Record the force on each of the spring
balances.
5 Using a protractor and trigonometry, determine the x and y components of the forces.
6 Now try starting with one force and predict the magnitude and direction of the other two forces required to keep the
ring from moving.
7 Apply the three forces to check if you were correct.
RECORD THIS…
Describe the relationship between the force components that resulted in the ring staying still.
Present a table of your results.
REFLECT ON THIS…
How are forces produced between objects and what effects do forces produce?
How did the ring move when all the forces were balanced?
How did the ring move when all the forces were unbalanced?
WS
2.1
The previous chapter developed the concepts and ideas needed to describe the motion
of a moving object. In this chapter you will investigate the forces that cause the motion.
FORCE
In simple terms, a force can be thought of as a push or a pull. You experience
forces all the time; they are fundamental to the nature of matter and the structure
of the universe.
In each of the situations depicted in Figure 4.1.1, forces are acting. Some are
applied directly to an object and some act on an object without touching it. A
force that acts directly on an object is called a contact force, because the object
experiences the force only while contact is maintained. A force that acts on an
object at a distance is called a non-contact force, or a force mediated by a field.
120
MODULE 2 | DYNAMICS
(a)
(b)
(d)
(c)
(e)
FIGURE 4.1.1 (a) At the moment of impact, both the tennis ball and the racquet strings are distorted
by the forces acting at this instant. (b) The rock climber is relying on the frictional force between the
rock face and his hands and shoes. (c) A continual force causes the clay to deform into the required
shape. (d) The gravitational force between the Earth and the Moon is responsible for two high tides
each day. (e) The magnet is suspended in mid-air because of magnetic forces.
Contact forces are the easiest to understand. They include the simple pushes and
pulls you experience every day, such as the forces that act between you and your
chair when you are sitting, or the forces that act between your hand and a ball when
you throw it. Friction and drag forces are also contact forces.
Forces mediated by a field occur when the object causing the push or pull is
physically separated from the object that experiences the force. These forces are
said to ‘act at a distance’. Gravitational, magnetic and electric forces are examples
of non-contact forces.
Force is measured using the SI unit called the newton, which has the symbol N.
This unit, which will be defined later in the chapter, honours Sir Isaac Newton
(1643–1727), who is one of the most important physicists in history and whose first
law is the subject of this section.
A force is a push or a pull and is measured in newtons (N). It is a vector, so it
requires a magnitude and a direction to describe it fully.
A force of one newton, 1 N, is approximately the force you have to exert when
holding a 100 g mass against the downwards pull of gravity. In everyday life this is
about the same as holding a small apple. If more than one force acts on an object at
the same time, the object behaves as if only one force—the vector sum of all the
forces—is
acting. The vector sum of the forces is called the resultant or net force,
Fnet. (Vectors are covered in detail in Chapter 2.)
NEWTON’S FIRST LAW
Many people mistakenly think that an object that is moving with a constant velocity
must have a force causing it to move. This section addresses this misunderstanding
and will enable you to understand how Newton’s first law applies to any situation
in which an object moves.
The net force acting on an
object experiencing a number of
forces acting simultaneously is
given by the vector sum of all the
individual forces:
Fnet = F1 + F 2 + ... + Fn
CHAPTER 4 | FORCES
121
Newton’s first law of motion is
commonly stated as follows:
Every object will remain in a
state of rest or uniform motion
in a straight line unless that state
is changed by the action of an
external force.
This law can be stated more
simply as:
An object will maintain a constant
velocity unless an unbalanced
external force acts on it.
Newton’s first law could be
rephrased to remove the term
‘unless’:
An object will not maintain a
constant velocity if an unbalanced
external force acts on it.
The simplified version of Newton’s first law in the box on the left needs to be
analysed in more detail by first examining some of the key terms used. The term
‘maintain a constant velocity’ implies that, if an object is moving and no external force
is acting on it, it will continue to move with a velocity that has the same magnitude and
direction. For example, if a car is moving at 12.0 m s−1 south, then some time later it
will still be moving at 12.0 m s−1 south (Figure 4.1.2). Similarly, if a car has a velocity
of 0 m s−1 (that is, it is not moving), then some time later it will still have a velocity
of 0 m s−1.
12.0 m s–1 south
FIGURE 4.1.2
12.0 m s–1 south
A car maintaining a constant velocity.
Newton’s first law is closely related to the concept of inertia; in fact it is sometimes
called the law of inertia. Inertia is the tendency of an object to maintain a constant
velocity. This tendency is related to the mass of an object: the greater the mass, the
harder it is to get it moving or to stop it from moving.
The term ‘unless’ is particularly important in Newton’s first law. Instead of
saying what must happen for the motion to be constant, it tells you what must
not happen.
In the rephrased version of the law shown on the left, ‘will not maintain a constant
velocity’ implies that the velocity will change. And a change in velocity means that
the object will accelerate.
The term ‘unbalanced’ implies that there must be a net force acting on the object.
If more than one force acts on an object but the forces are balanced, the object’s
velocity will remain constant. If the forces are unbalanced the velocity will change.
This is illustrated in Figure 4.1.3.
200 N south
200 N north
400 N south
200 N north
FIGURE 4.1.3 The forces on the top car are balanced, so it will maintain a constant velocity. The
forces on the bottom car are unbalanced: it has a net force in the forwards direction, so its velocity
will change.
The term ‘external’ in relation to a force implies that the force is not internal.
When forces are internal they have no effect on the motion of the object. For
example, if you are sitting in a car and push forwards on the steering wheel then the
car will not move forwards because of this force. In order for you to push forwards
on the steering wheel, you must push backwards on the seat. Both the steering wheel
and the seat are attached to the car, so there are two forces acting on the car that are
equal and in the opposite direction to each other, as shown in Figure 4.1.4. Because
internal forces must result in balanced forces on the object, they cannot change the
velocity of the object.
force of person
on seat/car
force of person
on wheel/car
FIGURE 4.1.4 This driver is applying internal forces on a car. These internal forces will balance and
cancel each other.
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PHYSICSFILE N
The effects of forces
Applying a force can cause an object to speed up, slow down, start moving, stop
moving, or change direction. The effect depends on the direction of the force in
relation to the direction of the velocity vector of the object experiencing the force.
The effect of external forces is summarised in Table 4.1.1.
TABLE 4.1.1 The effect of the application of a force, depending on the relationship between the
direction of the force and the velocity.
Relationship between velocity and
force
Effect of force
Force applied to object at rest.
Object starts moving.
Force applied in same direction
as velocity.
Magnitude of velocity increases (object
speeds up).
Force applied in opposite direction
to velocity.
Magnitude of velocity decreases (object
slows down).
Force applied perpendicular to velocity.
Direction of velocity changes (object
changes direction).
An unbalanced force applied to an object always changes the velocity of the object,
whether it is the magnitude of the velocity, the direction of the velocity, or both that
changes (as shown in Figure 4.1.3).
EQUILIBRIUM
Newton’s first law states that an object will continue with its motion unless acted
upon by an external unbalanced force. An object’s velocity will not change when the
forces acting on it are balanced. When the forces are balanced, the forces are said to
be in equilibrium.
An example of equilibrium occurs at the beginning of a game of tug-of-war
(Figure 4.1.5). Both teams apply a force in opposite directions but neither team
moves. Winning a tug-of-war game involves one team applying a greater force so
that there is a net force on the rope, causing the rope and teams to accelerate in the
winning team’s direction. When the rope and the teams are moving at a constant
velocity, then an equilibrium of forces exists once again.
The sum of all the forces acting on an object is commonly referred to as the
resultant force or net force, Fnet. An equilibrium occurs when the net force is zero.
A net force causes acceleration in one direction, but a zero net force causes no
acceleration of the object. This condition is thedefining aspect of an equilibrium of
forces. It can be expressed mathematically as Fnet = 0.
FIGURE 4.1.5 When a tug-of-war starts there is
an equilibrium of forces.
Equilibrium exists when the sum of all forces acting on an object result in a zero
net force acting on the object. Mathematically, this can be expressed as:
ΣF = 0
The symbol Σ (sigma) represents the sum. The equation ΣF = 0 means the sum
of all individual forces acting on an object. If there were four individual forces all
simultaneously acting on the same object, this equation would be rewritten as:
ΣF = F1 + F 2 + F 3 + F 4
CHAPTER 4 | FORCES
123
SKILLBUILDER
N
UNDERSTANDING MATHEMATICAL SYMBOLS
Part of the language of science is using symbols to represent quantities
or to give meanings. For example, the symbols <, >, ≤ and ≥ are known as
inequalities. Table 4.1.2 gives some examples of common mathematical
symbols used in science.
TABLE 4.1.2
Some mathematical symbols commonly used in science.
Symbol
Meaning
Example
<
less than
2<3
2 is less than 3
>
greater than
6>1
6 is greater than 1
≤
less than or equal to
2x ≤ 10
2x is less than or equal to 10
≥
greater than or equal to
3y ≥ 12
3y is greater than or equal to 12
square root
The square root of 4 is 2
4 =2
Δ
change in, or difference
between
Δt
≈
approximately equal to
π ≈ 3.14
∑
sum
4
∑i
i =1
∝
is proportional to
F ∝a
Explanation
change in t (time)
π is approximately equal to 3.14
The sum of consecutive integers
from 1 to 4, i.e.
1 + 2 + 3 + 4 = 10
Force is proportional to
acceleration
Worked example 4.1.1
CALCULATING NET FORCE AND EQUILIBRIUM IN ONE DIMENSION
Consider two forces acting on an object. A 10 N force acts on the object towards
the right, and a 5 N force acts on the object towards the left.
Calculate the net force acting on the object, and determine
what additional force
is required for the object to be in equilibrium. That is, ΣF = 0.
Thinking
Determine the individual forces acting
on the object.
Apply a sign convention to replace
directions.
Determine the net force acting on the
object.
Determine what additional force
is required for the object to be in
equilibrium
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MODULE 2 | DYNAMICS
Working
F1 = 10 N towards the right
F2 = 5 N towards the left
F1 = +10 N
F2 = −5 N
Fnet = F1 + F2
= 10 + (−5)
= 5 N towards the right
Equilibrium exists when ΣF = 0
Fnet = 5 N towards the right
For equilibrium to exist, a force of 5 N
towards the left would be required to
act on the object.
Worked example: Try yourself 4.1.1
CALCULATING NET FORCE AND EQUILIBRIUM IN ONE DIMENSION
Consider two forces acting on an object. A 20 N force acts on the object towards
the left, and a 23 N force acts on the object towards the right.
Calculate the net force acting on the object, and determine what additional force
is required for the object to be in equilibrium.
When you are working with vectors in two dimensions, you can describe the
vectors using their x and y components. For example, if a 10 N force is acting at a
45° angle, that vector could be represented as the sum of its x and y components, as
shown in Figure 4.1.6.
In Figure 4.1.6, if F = 10 N then the x and y components of the force are:
Fx = 10cos45°
Fx = 7.07 N
Fy = 10sin 45°
Fy = 7.07 N
→
F
→
→
→
Fy = Fsin 45º
45º
→
FIGURE 4.1.6 A two-dimensional vector can be
described by its x and y components. In this
figure, the x component of the force
is F cos 45°,
the y component of the force is F sin 45°.
Worked example 4.1.2
CALCULATING NET FORCE AND EQUILIBRIUM IN TWO DIMENSIONS
Consider three forces acting on an object: a 20 N upwards force, a 10 N downwards
force, and a 10 N force from left to right.
a Calculate the net force acting on the object
Thinking
Determine the individual
forces acting on the
object.
Working
F1 = 20 N upwards
F2 = 10 N downwards
F3 = 10 N right
Apply a sign convention
to replace directions.
Choose up and right to be positive directions. This
means down and left are negative.
Create a vector diagram
describing the net force
acting on the object.
→
F3
→
F2
→
→
F1
→
Fnet
→
Fy
F1
→
F3
Calculate the magnitude
of the net force using
Pythagoras’ theorem.
→
F2
→
Fnet
→
Fx
Fnet = F1 + F2 + F3
Fnet = 20 N (up) − 10 N (down) + 10 N (right)
= 10 N (up) + 10 N (right)
2
Fnet = 102 + 102
Fnet = 200
Fnet = 14.1 N
Determine the angle of
the net force.
tan θ = 10
10 = 1
State the net force.
= 45°
Fnet = 14.1 N at an angle of 45° in the right direction
from the horizontal
θ = tan−1 1
CHAPTER 4 | FORCES
125
b Determine what additional force is required for the object to be in equilibrium.
Thinking
Determine what
additional force is
required for the object to
be in equilibrium.
Working
Equilibrium exists when ∑ F = 0:
Fnet = 10 N upwards and 10 N towards the right
So for equilibrium to exist, an additional force of
10 N downwards and 10 N towards the left must act
on the object.
This vector is equal to the negative value of Fnet
calculated in part a.
→
F3
→
F2
→
F1
→
Fequilibrium
Worked example: Try yourself 4.1.2
CALCULATING NET FORCE AND EQUILIBRIUM IN TWO DIMENSIONS
Consider three forces acting on a single object: a 10 N downwards force, a 2 N
downwards force and a 5 N force from right to left.
a Calculate the net force acting on the object.
b Determine what additional force is required for the object to be in equilibrium.
Stating Newton’s first law in different ways
Ludwig Wittgenstein, an Austrian−British philosopher, suggested that ‘understanding means seeing that the same thing said in different ways is the same thing’.
To truly understand Newton’s first law, you should be able to state it in different ways
yet still recognise it as being consistent with Newton’s first law.
All of the following statements are consistent with Newton’s first law:
• An object will maintain a constant velocity unless an unbalanced, external force
acts on it.
• An object will continue with its motion unless an unbalanced, external force is
applied.
• An object will maintain a constant velocity when the forces acting on it are in
equilibrium.
• If the forces acting on an object are in equilibrium, the object will maintain a
constant velocity.
• An object will either remain at rest or continue with constant speed in a straight
line unless it is acted on by a net force.
• If an unbalanced external force is applied, then an object’s velocity will change.
• If a net force is applied (the forces are not in equilibrium), then the object’s
velocity will change.
• Constant velocity means net force is equal to zero.
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PHYSICS IN ACTION
N
Terminal velocity
You saw in Chapter 2 that without air resistance all objects
would accelerate towards the surface of Earth at a constant
rate of 9.8 m s−2. But air resistance exists on Earth, and it
increases as the speed of an object increases. Newton’s
first law can be used to explain how air resistance causes
a skydiver to reach a maximum vertical velocity, called the
terminal velocity.
As the skydiver begins falling, the only external force is
the weight force. The weight force causes the skydiver to
accelerate at 9.8 m s−2. But as their velocity increases, air
resistance pushes upwards. This reduces the magnitude of
the net force, which decreases the skydiver’s acceleration.
Eventually the air resistance becomes so great that it exactly
balances the weight force. According to Newton’s first law,
the skydiver will maintain a constant velocity when all the
external forces are balanced: this velocity is the terminal
velocity. Terminal velocity is different for different objects,
because air resistance depends on the object’s shape
and other factors. For example, terminal velocity for a
sheet of paper is much less than for a skydiver.
FIGURE 4.1.7 When a skydiver reaches terminal velocity, their
weight force is in balance with air resistance.
INERTIA
Inertia can be thought of as the resistance of an object to a change in motion. It
is related to the mass of the object. As the mass of the object increases, its inertia
increases, and therefore:
• it becomes harder to start it moving if it is stationary
• it becomes harder to stop it moving
• it becomes harder to change its direction of motion.
You can experience the effect of inertia when you push a trolley in a supermarket. If
the trolley is empty it is easy to start pushing it, and to pull it to a stop when it is already
moving. It is also easy to change its direction of motion. But when the trolley is filled
with heavy groceries it becomes more difficult—that is, it requires more force—to make
the trolley start moving when it is at rest, and it becomes more difficult to pull it to a
stop if it is already moving. It also requires more force to change the trolley’s direction.
It is important to note that the effects of inertia are independent of gravity.
Because inertia depends on mass, and weight force due to gravity also depends on
mass, it is a common misunderstanding to think that the effects of inertia only occur
in the presence of gravity. So even if your supermarket was in deep space, it would
be just as difficult to change the state of motion of the trolley.
Newton’s first law and inertia
The connection between Newton’s first law and inertia is very close. Because of
inertia, an object will continue with a constant motion unless a net force acts on it.
You experience the connection between Newton’s first law and inertia when you
are standing in a train that is initially at rest but then starts moving forwards. If you
are not holding on to anything, you may stumble backwards as though you have
been pushed backwards. What has happened?
You have not been pushed backwards. Because you have inertia, your mass
resists the change in motion when the train starts moving. According to Newton’s
first law, your body is simply maintaining its original state of being motionless until
an unbalanced force acts to accelerate it. When the train later comes to a sudden
stop, your body again resists the change by continuing to move forwards until an
unbalanced force acts to bring it to a stop.
CHAPTER 4 | FORCES
127
4.1 Review
SUMMARY
• Newton’s first law can be written in many ways;
for example:
• A force is a push or a pull. Some forces act by
contact, while others can act at a distance.
-- An object will continue with a constant
velocity unless an unbalanced force causes its
velocity to change.
• Force is a vector quantity, and its SI unit is the
newton (N).
• When an object experiences more than one force
acting simultaneously, the net force is given by the
vector sum of the individual forces.
-- Non-zero net forces cause acceleration.
• Inertia is the tendency of an object to resist a
change in motion.
• Equilibrium exists when the sum of all forces
acting on an object results in a zero net force.
• Inertia is related to mass; an object with a larger
mass will have a larger inertia.
KEY QUESTIONS
1
A student observes a box sliding across a surface and
slowing down to a stop. From this observation, what
can the student conclude about the forces acting on
the box?
2
A car changes its direction as it turns a bend in the
road while maintaining its speed of 80 km h−1. From
this, what can you conclude?
3
4
128
A bowling ball rolls forwards along a smooth wooden
floor at a constant velocity. Ignoring the effects of friction
and air resistance, which of the following statements,
relating to the force acting on the ball, is correct?
A There is a net force acting forwards to maintain the
velocity of the ball.
B There is not an unbalanced force acting on the ball.
C The forwards force acting on the ball is balanced by
the friction that opposes the motion.
D It is not possible to conclude anything about the
forces on the ball.
If a person is standing up in a moving bus that stops
suddenly, the person will tend to fall forwards. Has a
force acted to push them forwards? Use Newton’s first
law to explain what is happening.
MODULE 2 | DYNAMICS
5
Passengers on commercial flights must be seated and
have their seatbelts done up when the aeroplane is
coming in to land. What would happen to a person
who was standing in the aisle as the aeroplane
travelled along the runway during landing?
6
A magician performs a trick in which a cloth is pulled
quickly from under a glass filled with water without
causing the glass to fall over or the water to spill out.
a Explain the physics underlying this trick.
b Does using a full glass make the trick easier or
more difficult? Explain.
7
When flying at constant speed at a constant altitude,
a light aircraft has a weight of 50 kN down, and the
thrust produced by its engines is 12 kN to the east.
What is the lift force acting on the wings of the plane?
And what is the magnitude and direction of the drag
force?
8
If an object experiences a force of 20 N towards the
right and a force of 15 N towards the left, what is
the net force acting on the object? Determine what
additional force is required for the object to be in
equilibrium.
4.2 Newton’s second law
Many people mistakenly believe that heavy objects will fall faster than lighter objects.
This is because air resistance has a significant effect on larger, lighter objects.
However, even when air resistance is taken into account, the misconception that
larger masses should fall faster than lighter masses persists.
Newton’s second law makes the quantitative connection between force, mass
and acceleration. It helps to resolve the misconception about the time taken for
objects with different masses to fall to the ground.
Figure 4.2.1 depicts a famous experiment showing that objects fall together
when the effect of air resistance is removed. An internet search for ‘hammer and
feather on the Moon’ will enable you to view a video of Apollo astronaut David
Scott’s 1971 experiment (see the PhysicsFile at the end of Chapter 2). Although
the video is fuzzy, you can see that both objects accelerate at the same rate and
hit the ground at the same time. You can also find a video of Professor Brian Cox
performing the same experiment in a huge vacuum chamber used by NASA for
testing spacecraft, using feathers and a bowling ball.
NEWTON’S SECOND LAW
FIGURE 4.2.1 An artist’s image of the famous
hammer and feather experiment conducted on
the Moon.
GO TO ➤ PhysicsFile, page 85
Newton’s second law of motion states:
The acceleration of an object is directly proportional to the net force on the
object and inversely proportional to the mass of the object:
F = ma
where: a is the acceleration of an object (in m s−2)
F is the force applied to the object (in N)
m is the mass of the object (in kg)
By definition, 1 N is the force needed to accelerate a mass of 1 kg at 1 m s−2. It is
therefore equal to 1 kg m s−2.
One of the implications of Newton’s second law is that, for a given mass,
a greater acceleration is achieved by applying a greater force. This is shown in
Figure 4.2.2. Doubling the applied force will double the acceleration of the object.
In other words, acceleration is proportional to the net force applied.
F1⃗
a1⃗
mass
F2⃗
a2⃗
mass
FIGURE 4.2.2 Given
the same mass, a larger force will result in a larger acceleration. If the force is
doubled, then the acceleration is also doubled.
Notice also in Figure 4.2.2 that the acceleration of the object is in the same
direction as the net force applied to it.
Newton’s second law also explains how acceleration is affected by the mass of an
object. For a given force, the acceleration of an object will decrease with increased
mass. In other words, acceleration is inversely proportional to the mass of an object.
This is shown in Figure 4.2.3.
Applying Newton’s second law
In many practical situations, such as designing lifts for skyscrapers or seatbelts for
cars, it is important to calculate the force acting
a mass that is accelerating (or
on
decelerating) at a certain rate. The equation F = ma enables us to do this easily.
F⃗
m1
a1⃗
F⃗
m2
a2⃗
FIGURE 4.2.3 Given the same force, a larger
mass will result in a lower acceleration. If the
mass is doubled, then the acceleration is halved.
CHAPTER 4 | FORCES
129
Worked example 4.2.1
CALCULATING THE FORCE THAT CAUSES AN ACCELERATION
Calculate the net force on a 5.50 kg mass that is accelerating at 3.75 m s−2 west.
Thinking
Working
Ensure that the variables are in their
standard units.
m = 5.50 kg
a = 3.75 m s−2 west
F = ma
Apply the equation for force from
Newton’s second law.
= 5.50 × 3.75
= 20.6 N
F = 20.6 N west
Give the direction of the net force,
which is always the same as the
direction of the acceleration.
Worked example: Try yourself 4.2.1
CALCULATING THE FORCE THAT CAUSES AN ACCELERATION
Calculate the net force on a 75.8 kg runner who is accelerating at 4.05 m s−2 south.
GO TO ➤ Section 2.1, page 50
Often more than one force will act on an object at a particular time. The overall
effect of the forces depends on the direction of each of the forces. For example,
some forces may reinforce each other, but others may oppose each other. When
using Newton’s second law it is important to use the net, or resultant, force in the
calculation. Because forces are vectors, they can be added or combined using the
techniques discussed in Chapter 2. Consider the following worked examples.
Worked example 4.2.2
CALCULATING THE ACCELERATION OF AN OBJECT WITH MORE THAN ONE
FORCE ACTING ON IT
A swimmer whose mass is 75 kg applies a forwards force of 50 N to the water
as he starts a lap. The water opposes his efforts to accelerate with a backwards
drag force of 20 N.
What is the swimmer’s initial horizontal acceleration?
Thinking
Determine the individual forces
acting on the swimmer, and apply
the vector sign convention.
Determine the net force acting on
the swimmer.
Working
F1 = 50 N forwards
= +50 N
F2 = 20 N backwards
= −20 N
Fnet = F1 + F2
= 50 + (−20)
= + 30 N or 30 N forwards
⃗
Fapplied
+
⃗
Fdrag
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MODULE 2 | DYNAMICS
50 N
20 N
ΣF ⃗
30 N
Use Newton’s second law to
determine acceleration.
a=
=
Fnet
m
30
75
= 0.40 m s−2 forwards
Worked example: Try yourself 4.2.2
CALCULATING THE ACCELERATION OF AN OBJECT WITH MORE THAN ONE
FORCE ACTING ON IT
A car with a mass of 900 kg applies a driving force of 3000 N forwards as it
starts moving. Friction and air resistance oppose the motion of the car with a
force of 750 N backwards.
What is the car’s initial acceleration?
Worked example 4.2.3
CALCULATING THE ACCELERATION OF AN OBJECT WITH A TWO-DIMENSIONAL
FORCE ACTING ON IT
A 200 N force acts at an angle of 45° to the x direction on an object with a mass
of 100 kg. A second force of 100 N acts on the same object at an angle of 30° to
the x direction.
→
F2
30º
y
x
→
F1
45º
What is the net force and initial acceleration acting on the object in the x
direction?
Thinking
Determine the horizontal (x)
components of the forces acting on
the object.
Working
F1 = 200 N at 45°
F = 100 N at 30°
2
F1( x ) = 200 cos 45°
= +141.4 N in the x direction
F2( x ) = 100 cos 30°
Determine the net force acting on the
object in the x direction.
= +86.6 N in the x direction
Fnet = F1( x ) + F2( x )
= 141.4 + 86.6
= +228 N in the x direction
Use Newton’s second law to determine
acceleration.
a=
Fnet
m
=
228
100
= +2.28 m s−2 in the x direction
PHYSICSFILE N
The weight force
The mass of an object is different to its
weight. Mass is a scalar quantity and is
measured in kilograms (kg). Weight is
the
force on an object due to gravity,
Fg. Like all forces, weight is a vector
quantity and is measured
in newtons
(N). The equation for Fg has a similar
form to Newton’s second law, F = ma.
F = mg
where:
m is the mass of the
object (kg)
g is the gravitational field
strength (N kg−1)
The gravitational field strength is
equivalent to the acceleration of an
object due to gravity, so its unit can
also be written as m s−2.
CHAPTER 4 | FORCES
131
Worked example: Try yourself 4.2.3
CALCULATING THE ACCELERATION OF AN OBJECT WITH A TWO-DIMENSIONAL
FORCE ACTING ON IT
A 150 N force acts at an angle of 60° to the x direction on an object with a mass
of 75 kg. A second force of 80 N acts on the same object at an angle of 15° to
the x direction.
→
F2
15º
y
→
F1
x
60º
PA
2.3
PA
2.3
What is the net force and initial acceleration acting on the object in the y
direction?
PHYSICSFILE ICT
Maximising acceleration
Dragster race cars are designed to achieve the maximum possible acceleration in order
to win a race in a straight line over a short distance. According to Newton’s second law,
acceleration is increased by increasing the applied force and by reducing the mass of
the object. For this reason, dragster race cars are designed with very powerful engines
that produce an enormous forwards force and an aerodynamic shape to minimise air
resistance. There is not much else to the car, so this helps to minimise the mass.
Newton’s second law also helps you understand why a motorcycle can accelerate
FIGURE 4.2.4 The lower mass of motorcycles enables
from the lights at a greater rate than a car or a truck. While the engines in a car or
truck are usually more powerful than a motorcycle engine, the motorcycle has much them to accelerate faster than cars and trucks, even
though their engines are relatively small.
less mass, which allows for greater acceleration.
THE FEATHER AND HAMMER EXPERIMENT
The experiment on the Moon
m2
m1
F1⃗
ag⃗
F2⃗
ag⃗
FIGURE 4.2.5 Both objects experience the same
acceleration, so the larger mass must have a
larger force acting on it. If the mass is doubled,
then the force is doubled.
When two objects with different masses fall under the influence of a gravitational
field, in the absence of any other force such as air resistance, they will both fall at
the same rate. That is, their accelerations will be the same. They will cover the same
distance in the same time and will hit the ground at the same time if dropped from
the same height. This happens on the Moon because there is no atmosphere, and
therefore no air resistance.
Many people think that a gravitational field applies the same force to all objects.
In fact, the gravitational force is larger on objects with more mass, and smaller on
objects with less mass. A larger mass also has more inertia, so it requires a greater
force to achieve the same acceleration (Figure 4.2.5).
The experiment on the Earth
When a feather falls through the air on Earth, its acceleration is far less than the
acceleration of a hammer falling from the same height. From the previous section
you will know that the hammer and the feather have gravitational forces acting on
them that are proportional to their mass, so they should accelerate at the same rate.
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MODULE 2 | DYNAMICS
They do not do this because of the force of air resistance. Recall that Newton’s
second law says the acceleration is proportional to the net force acting on an object,
which means you must consider all the forces acting on an object to determine
the acceleration.
Air resistance is a force that results from air molecules colliding with the object.
The faster an object moves, the greater the air resistance. Air resistance also
increases with increasing surface area perpendicular to the direction of motion, and
with the roughness of the surface (represented by the drag coefficient). This force,
which acts in the opposite direction to the motion of the object, is significant when
compared with the weight of the feather, but insignificant when compared with the
weight of the hammer (Figure 4.2.6).
Because an average feather has a very small mass (less than 0.5 g), the gravitational
force acting on it is very small. It also has a relatively large surface area and a large
drag coefficient, so the air resistance when it starts falling is large relative to the
gravitational force acting on it. As soon as the feather is released its acceleration
decreases quickly, and it reaches its terminal velocity after falling only about a metre.
This velocity is typically about 1 m s−1.
On the other hand, a hammer has a large mass (about 700 g), a small surface
area (because it falls with the heavy end down), and a small coefficient of drag. As
a result the air resistance when it starts falling is small compared to the gravitational
force acting on it, and it does not reach its terminal velocity until it has fallen
hundreds of metres. This velocity is about 50 m s−1.
F⃗AR2
F⃗AR1
m2
m1
F⃗1
a⃗g
a⃗g
F⃗2
FIGURE 4.2.6 The net force on a falling object on Earth is a result of the gravitational force acting
downwards and air resistance acting upwards. For an object with a small mass (m1), air resistance
has a large effect on the initial acceleration because the gravitational force acting on the object is
relatively small. For a heavy object (m2), air resistance has little effect on the initial acceleration
because the gravitational force acting on the object is relatively large.
CHAPTER 4 | FORCES
133
4.2 Review
SUMMARY
• Newton’s second law states:
• Different magnitudes of forces due to gravity
act on different masses to cause the same
acceleration.
The acceleration of an object is directly
proportional to the net force on the object and
inversely proportional to the mass of the object.
• Air resistance is a force that acts to decrease the
acceleration of objects moving through air.
relationship can be described by the formula
• This
F = ma.
KEY QUESTIONS
Use g = 9.8 m s−2 when answering the following questions.
1
Calculate the acceleration of a 23.9 kg mass when a
net force of 158 N north acts on it.
2
Calculate the acceleration of a 45.0 kg mass that has a
net force of 441 N downwards acting on it.
3
Calculate the acceleration of a 90.0 kg mass that has a
net force of 882 N downwards acting on it.
4
Calculate the mass of a train if it accelerates at
7.20 m s−2 north when a net force of 565 000 N north
acts on it. Give your answer to three significant figures.
5
A model yacht with a mass of 15 kg is sailing north
across a lake. A southerly breeze provides a constant
driving force of 10 N north, and the drag forces total
8.5 N south.
a Calculate the weight of the model.
b Find the net force acting on the model when it is
sailing.
134
MODULE 2 | DYNAMICS
c Calculate the magnitude of the acceleration of the
model.
d The breeze drops and the yacht now travels at a
constant speed north, with total drag forces of 7.2 N
south. What is the force of the breeze on the model
now?
6
An empty truck of mass 2000 kg has a maximum
acceleration of 2.0 m s−2. How many 300 kg boxes of
goods would the truck be carrying if its maximum
acceleration was 1.25 m s−2?
4.3 Newton’s third law
Newton’s first two laws of motion describe the motion of an object resulting from the
forces that act on that object. Newton’s third law, concerning ‘action and reaction’,
is easily stated and is well known, but it is often misunderstood and misused. It is a
very important law in physics because it helps us understand the origin and nature
of forces. Newton’s third law is explored in detail in this section.
NEWTON’S THIRD LAW
Newton realised that all forces exist in pairs, and that each force in the pair acts
on a different object. Consider a hammer hitting a nail on the head, as shown in
Figure 4.3.1. Both the hammer and the nail experience forces during this interaction.
The nail experiences a downwards force as the hammer hits it. When the nail is
hit it moves a distance into the wood. As it hits the nail, the hammer experiences
an upwards force that causes the hammer to stop. These forces are known as an
action–reaction pair, and are shown in Figure 4.3.2.
(a)
(b)
FIGURE 4.3.1 A hammer hitting a nail is a
good example of an action–reaction pair and
Newton’s third law.
force that
nail exerts
on hammer
force that
hammer exerts
on nail
force that
nail exerts
on hammer
force that
hammer exerts
on nail
(a)
force that foot
exerts on ball
FIGURE 4.3.2 (a) As the hammer gently taps the nail, both the hammer and the nail experience small
forces. (b) When the hammer hits the nail harder, both the
hammer and thenail experience larger
forces. In both cases these forces can be designated by Fby hammer on nail and Fby nail on hammer.
Regardless of whether the hammer exerts a small or a large force on the nail,
the nail will exert exactly the same magnitude of force on the hammer, but in the
opposite direction.
Newton’s third law of motion states:
For every action (force), there is an equal and opposite reaction (force).
Newton’s third law means that when object A exerts a force on object B, object
B will exert an equal and opposite force on object A. It is important to recognise that
the action force and the reaction force in Newton’s third law act on different objects
and so should never be added together; their effect will only be on the object on
which they act. Newton’s third law applies not only to forces between objects which
are in direct contact, but also to non-contact forces such as gravitational force.
The main misunderstanding that arises when considering Newton’s third law is
the belief that, if a large mass collides with a smaller mass, then the larger object exerts
a larger force and the smaller object exerts a smaller force. This is not true. If you
witnessed the collision between the car and the bus in Figure 4.3.3b, you would see the
car undergoing a large deceleration while the bus undergoes only a small acceleration.
From Newton’s second law, you know that the same force acting on a larger
mass will result in a smaller acceleration. This is the effect seen in the situation of the
car colliding with the bus. Because of the car’s small mass, the force acting on the car
will cause the car to undergo a large deceleration. The occupants may be seriously
injured as a result of this. The force acting on the bus is equal in size, but is acting
on a much larger mass. As a result, the bus will have a relatively small acceleration
and the occupants will not be as seriously affected.
force that ball
exerts on foot
(b)
force that bus
exerts on car
force that car
exerts on bus
(c)
gravitational force
that Earth exerts
on brick (Fg⃗ )
gravitational force
that brick exerts
on Earth
FIGURE 4.3.3
Some action–reaction pairs.
CHAPTER 4 | FORCES
135
Identifying the action and reaction forces
When analysing a situation to determine the action and reaction forces according to
Newton’s third law, it is helpful to be able to label the force vectors systematically. A
good strategy for labelling force vectors is to use a subscript consisting of the word
‘by’ and the object applying the force, and then the word ‘on’ and the object that is
acted on by the force. Once you are familiar with this system, you can just use letters
or numbers to represent each object.
The equal and opposite force is then labelled with a subscript that has the
objects in reverse. For example, the action and reaction force vector arrows shown
in Figure 4.3.3 can be labelled as shown in Table 4.3.1.
TABLE 4.3.1
Labels of action and reaction force vectors in Figure 4.3.3.
Action vector
or FFB
Reaction vector
Fby ball on foot or FBF
Fby car on bus or FCB
Fby bus on car or FBC
Fby brick on Earth or FBE
Fby Earth on brick or FEB
Fby foot on ball
It does not matter which force is considered the action force and which is considered
the reaction force. They are always equal in magnitude and opposite in direction.
Newton’s
third law can be expressed mathematically as:
F AB = −F BA
where −FBA is the action force applied by object A on object B, and is the equal
and opposite reaction force applied by object B on object A.
GO TO ➤ Section 3.2, page 106
This is similar to the vector rule described in Chapter 3.
Worked example 4.3.1
PHYSICSFILE ICT
Combining Newton’s second
and third laws
You can easily observe the effect of
Newton’s second and third laws in the
classroom if you have two dynamics
carts with wheels that are free to roll
on a smooth surface such as a bench
or desk. If the two carts are placed
in contact with each other and the
plunger is activated on one of the
carts, the carts will roll away from each
other. This is because of the action
and reaction force pair described by
Newton’s third law.
If the two carts have similar masses,
they will accelerate apart at a similar
rate. If one cart is heavier than the
other, the lighter cart will accelerate
at a greater rate. This is because the
forces acting on both carts are equal
in magnitude and so, according to
Newton’s second law, the smaller mass
will experience a greater acceleration.
136
MODULE 2 | DYNAMICS
APPLYING NEWTON’S THIRD LAW
In the diagram below, a table tennis bat is in contact with a ball, and the action
force vector is shown.
a Identify the action and reaction forces using the system Fby ... on ...
Thinking
Working
Identify the two objects involved in the action–
reaction pair.
The bat and the ball.
Identify which object is applying the force and
which object is experiencing the force, for the
force vector shown.
The action force vector shown
is a force by the bat on the ball.
Use the system of labelling action and
reaction forces to label the action and
reaction forces.
Action force: Fby bat on ball
Reaction force: Fby ball on bat
b Draw the reaction force on the diagram, showing its size and location, and
label both forces.
Thinking
Working
Copy the diagram into your
workbook.
Use a ruler to measure the length
of the action force and construct
a vector arrow in the opposite
direction with its tail on the point of
application of the reaction force.
forceby ball on bat
forceby bat on ball
Label the forces.
Worked example: Try yourself 4.3.1
APPLYING NEWTON’S THIRD LAW
In the diagram below, a bowling ball is resting on the floor, and the action force
vector is shown.
a Identify the action and reaction forces using the system Fby ... on ...
b Draw the reaction force on the diagram, showing its size and location, and
label both forces.
NEWTON’S THIRD LAW AND MOTION
Newton’s third law explains how you are able to move around. In fact, Newton’s third
law is needed to explain all motion. Consider walking.Your leg pushes backwards on
the ground with each step. An action force is applied by your shoe on the ground.
As shown in Figure 4.3.4, one component of the force acts downwards and another
component pushes backwards horizontally along the surface of the ground.
The force is transmitted because there is friction between your shoe and the
ground. In response, the ground then pushes forwards on your foot. This forwards
component of the reaction force enables you to move forwards. In other words, it
is the ground pushing forwards on you that moves you forwards.
It is important to
F
=
ma
, the net force
remember
that
in
the
equation
for
Newton’s
second
law,
net
Fnet is the sum of the forces acting on the object. This does not include forces that
are exerted by the object on other objects. When you push back on the ground, this
force is acting on the ground and may affect the ground’s motion. If the ground is
firm this effect is usually not noticed, but if you run along a sandy beach, the sand
is pushed back by your shoes.
The act of walking relies on there being some friction between your shoe and
the ground. Without it there is no grip and it is impossible to supply the action
force to the ground. Consequently, the ground cannot supply the reaction force
needed to enable forwards motion. Walking on smooth ice is a good example of this.
Mountaineers attach crampons (metal plates with spikes) to the soles of their boots
to gain more grip on ice. The effects of friction on Newton's laws is discussed in
more detail in Chapter 5.
F ⃗ (reaction)
θ
F ⃗ (action)
⃗
Fforwards
⃗
Fup
θ
θ
F ⃗ (reaction)
FIGURE 4.3.4 Walking relies on an action–
reaction pair in which the shoe pushes down
and backwards with an action force. In
response, the ground pushes upwards and
forwards on the shoe.
CHAPTER 4 | FORCES
137
All motion can be explained in terms of action and reaction force pairs.Table 4.3.2
gives some examples of the action and reaction pairs in familiar motions.
TABLE 4.3.2
Action and reaction force pairs are responsible for all types of motion.
Motion
Action force
Reaction force
swimming
hand pushes back on water
water pushes forwards on hand
jumping
legs push down on Earth
Earth pushes up on legs
cycling
tyre pushes back on ground
ground pushes forwards on tyre
launching a rocket
rocket pushes gases downwards
gases push upwards on rocket
typing
finger pushes down on key
key pushes up on finger
THE NORMAL FORCE
Fg⃗
FIGURE 4.3.5 When the bin is in mid-air there is
an unbalanced gravitational force acting on it, so
it accelerates downwards.
When an object such as a bin is allowed to fall under the influence of gravity, it is
easy to see the effect of the gravitational force due to gravity. The action force is the
gravitational force of the Earth on the bin, so the net force on the bin in Figure 4.3.5
(ignoring air resistance) is equal to the gravitational force, and the bin therefore
accelerates at 9.8 m s−2.
When
the bin is at rest on a table, as shown in Figure 4.3.6a, the gravitational
force F g = mg is still acting on it. But the bin is at rest, so there must be another force
acting upwards to balance the gravitational force. This upwards force is provided
by the table. Because gravity pulls down on the mass of the bin, the bottom of the
bin pushes down on the surface of the table, and the table provides a reaction force
on the bin that is equal and opposite, so it pushes upwards on the bin as shown in
Figure 4.3.6b.
(a)
(b)
gravitational force
Fg⃗ = Earth exerts on bin
gravitational force
FG⃗ = bin exerts on Earth
FN⃗ = force on bin
by table
force on table
FT⃗ = by bin
FIGURE 4.3.6 (a) Action–reaction gravitational forces between the bin and the Earth. (b) Action–
reaction contact forces between the bin and the table.
FN⃗
Fg⃗
FIGURE 4.3.7 The effect of the two forces acting
on the bin. These are not an action–reaction
force pair, even though they are equal in
magnitude and opposite in direction.
138
MODULE 2 | DYNAMICS
The magnitude and direction of the gravitational force on the bin is equivalent
to the magnitude and direction of the force applied by the bin to the table. Therefore
the gravitational force on the bin is balanced by the upwards contact reaction force
applied by the table on the bin.
It is important to note that these two forces are not the pair of forces described in
Newton’s third law (Figure 4.3.7). This is because the two forces are both acting on
the bin, and no pair of Newton’s third law force pairs acts on the same object. The
contact force applied by a surface that is perpendicular to another surface is called
the normal reaction force.
It is often abbreviated to normal force and is usually
represented by the symbol FN.
When you consider only the forces acting on the bin, you are left with the
gravitational force on the bin and the normal force on the bin by the table. These
two forces come from two separate pairs of action–reaction forces.
The weight force and the normal force do not make up an action–reaction pair
under Newton’s third law, because they both act on the same object.
THE INCLINED PLANE
In the example of the bin and the table, the table’s surface is horizontal. An object
could be placed on a surface that is tilted at an angle θ to the horizontal,
as shown in
Figure 4.3.8. In this case the weight force remains the same: FN = mg downwards.
However, the normal force continues to act at right angles to the surface and will
change in magnitude, getting smaller as the angle increases. The magnitude of the
normal force is equal in size but opposite in direction to the component
weight
of the
force that acts at right angles to the surface. So the normal force is FN = mg cosθ .
The component of the weight force that acts parallel to the surface will cause
the mass to slide down the incline. The motion of the object will then be affected
by friction, if it is present.
The component of the weight force that acts along the
surface is given by F = mg sin θ .
(a)
F⃗N
F⃗g
(b)
ΣF⃗ = Fg⃗ + F⃗N
=0
F⃗N
ΣF⃗ = Fg⃗ + F⃗N
FN⃗
F⃗g
F⃗N θ
body remains
at rest
θ
ΣF⃗
F⃗g
F⃗g
FIGURE 4.3.8 (a) Where the surface is perpendicular to the gravitational force, the normal force acts
directly upwards. (b) On an inclined plane, FN is at an angle to Fg and is given by FN = mg cosθ . If no
friction acts, the force that causes the object to accelerate down the plane is FN = mg sinθ .
PHYSICS IN ACTION
WS
2.2
WS
2.3
PA
2.1
PA
2.1
CCT
Newton’s laws in space
On Earth we are used to the effects of Newton’s three laws and usually
do not notice them. But astronauts at the International Space Station
(ISS) feel like they are weightless, so the effects of Newton’s three laws
can be more obvious.
Newton’s first law is apparent when the ISS needs to change its orbit. If
an astronaut is not holding onto something, they will stay floating in the
same position while the ISS moves around them.
If an astronaut throws an object to a crewmate, they need to take care
about how much force they put behind it because of Newton’s second
law. On Earth we’re used to allowing for the effect of gravity when we
throw things, but for astronauts in the ISS the direction of motion of an
object does not change once they have thrown it. The object might miss
the crewmate altogether and hit something important.
And each time an astronaut needs to push or press down on something,
such as pushing closed a locker door or pressing keys on their computer,
Newton’s third law means that there is an equal and opposite force
pushing back. Astronauts often need to strap themselves into their desk
so they don’t float away while they are working.
FIGURE 4.3.9 Astronaut Karen Nyberg aboard the International
Space Station.
CHAPTER 4 | FORCES
139
4.3 Review
SUMMARY
• When an object exerts a downwards force on a
surface, an equal and opposite reaction force is
exerted upwards by the surface on the object. This
is called the normal force.
• Newton’s third law states:
-- For every action (force), there is an equal and
opposite reaction (force).
• If the action force is labelled systematically, the
reaction force can be described by reversing the
label of the action force.
• If an object is on an inclined plane, the normal
force acting on the object from the plane
decreases as the angle of the plane increases.
• The action and reaction forces are equal and
opposite even when the masses of the colliding
objects are very different.
• The individual forces making up an action–reaction
pair act on different masses to cause different
accelerations according to Newton’s second law.
KEY QUESTIONS
1
What forces act on a hammer and a nail when a heavy
hammer hits a small nail?
2
In the figure below, an astronaut is orbiting the Earth,
and one of the forces acting on him is shown by the
red arrow.
4
A 70 kg angler is fishing in a 40 kg dinghy at rest on a
still lake when suddenly he is attacked by a swarm of
wasps. To escape, he leaps into the water and exerts a
horizontal force of 140 N north on the boat.
a What force does the boat exert on the fisherman?
b With what acceleration will the boat move initially?
c If the force on the fisherman lasted for 0.50 s,
determine the initial speed attained by both the
man and the boat.
5
An astronaut becomes untethered during a space-walk
and drifts away from the spacecraft. To get back to the
ship, she decides to throw her tool kit away. In which
direction should she throw the tool kit?
6
Two students, James and Tania, are discussing the
forces acting on a lunchbox that is sitting on the
laboratory bench. James states that the weight force
and the normal force are acting on the lunchbox and
that since these forces are equal in magnitude but
opposite in direction, they comprise a Newton’s third
law action–reaction pair. Tania disagrees, saying that
these forces are not an action–reaction pair. Who is
correct and why?
7
A 10 kg object is resting on an inclined plane at an
angle of 60° to the horizontal.
What is the force on the
objectdue to gravity Fg, and the perpendicular normal
force FN from the inclined plane on the object?
a Name the given force using the system Fby ... on ....
b Name the reaction force using the system Fby ... on ....
3
140
A swimmer completes a training drill in which she
doesn’t use her legs to kick, but uses only her arms to
move down the pool. What force causes her to move
forwards down the pool?
MODULE 2 | DYNAMICS
Chapter review
KEY TERMS
Newton’s third law
normal reaction force
terminal velocity
weight
mass
net force
newton
Newton’s first law
Newton’s second law
contact force
equilibrium
force
force mediated by a field
inertia
KEY QUESTIONS
1
Which of the following are examples of contact forces?
A two billiard balls colliding
B the electrostatic force between two charged
particles
C a bus colliding with a car
D the magnetic force between two fridge magnets
2
A student is travelling to school on a train. When
the train starts moving, she notices that passengers
tend to lurch towards the back of the train before
regaining their balance. Has a force acted to push the
passengers backwards? Explain your answer.
3
A bowling ball rolls along a smooth wooden floor at
constant velocity. Which of the following diagrams
correctly indicates the horizontal forces acting on
the ball as it rolls towards the right? (The weight and
normal force can been ignored.)
A
C
4
5
6
B
F⃗
F⃗
F⃗
D
F⃗
7
Consider three forces acting on a single object: a 20 N
upwards force, a 10 N downwards force, and a 10 N
force from left to right. Sketch:
a the vector diagram of the three forces and the
resultant force
b the force required for the object to be in
equilibrium.
8
If two equal masses experience the same force, which
of the following describes their accelerations?
A equal and opposite
B equal and in the same direction
C different and opposite
D different and in the same direction
9
Calculate the mass of an object if it accelerates at
9.20 m s−2 east when a force of 352 N east acts on it.
10 Calculate the acceleration of a 60.9 g golf ball when a
net force of 95.0 N south acts on it.
11 Calculate the acceleration of a 657 kg motorbike when
a net force of 3550 N north acts on it.
F⃗
F⃗
A force of 10 N acts from left to right on an object, and
a force of 5 N simultaneously acts from right to left on
the same object.
a What is the net force acting on the object?
b Are the forces in equilibrium?
Newton’s first law states that an object will maintain a
constant velocity unless an unbalanced, external force
acts on it. What distinguishes an external force from an
internal force?
12 A 150 N force acts at a 45° angle to the x direction on
an object with a mass of 10 kg. A second force of 15 N
acts on the same object at an angle of 30° to the x
direction. Using the diagram below, calculate the net
force and initial acceleration acting on the object in the
x direction.
→
F2
30º
→
F1
y
x
45º
What are the horizontal and vertical components
of a force of 50 N acting on an object at an angle of
45° upwards from the positive x direction?
CHAPTER 4 | FORCES
141
CHAPTER REVIEW CONTINUED
13 A 10 N force acts at a 60° angle to the x direction on
an object with a mass of 75 kg. A second force of 40 N
acts on the same object at an angle of 15° to the x
direction. What is the net force and initial acceleration
acting on the object in the y direction?
→
F2
15º
y
→
F1
x
60º
14 Compare the acceleration of a motorcycle to the
acceleration of a bus. Which of the two vehicles will
accelerate at a greater rate, and why?
15 An object of mass 7.0 kg rests on an inclined plane at
an angle of 65°. Calculate the magnitude of the normal
force applied by the inclined plane to the object.
16 The thrust force of a rocket with a mass of 50 000 kg is
1000 kN. Calculate the magnitude of its acceleration.
142
MODULE 2 | DYNAMICS
17 Which of the following is a correct version of Newton’s
third law?
A If a swimmer’s hand pushes back on the water, the
water does not push back on the swimmer’s hand.
B If a bicycle tyre pushes backwards on the ground,
the person riding the bike also pushes backwards
on the ground.
C If a person jumps upwards, their legs push down on
the Earth and the Earth pushes up on their legs.
18 When an inflated balloon is released it will fly around
the room. What is the force that causes the balloon
to move?
19 Determine the reaction force involved when a ball is hit
with a racquet with a force of 100 N west.
20 A 65.0 kg skateboarder is standing on a 3.50 kg
skateboard at rest when he steps off the board and
exerts a horizontal force of 75.0 N south on the board.
What force does the board exert on the skateboarder?
21 After completing the activity on page 120, reflect on
the inquiry question:
How are forces produced between objects and what
effects do forces produce? In your response, discuss
the concepts of equilibrium and net force.
CHAPTER
Forces, acceleration
and energy
Energy conversion is a common thread in many daily activities, as well as some
more extreme activities. Your body’s energy stores are used constantly, especially
when you are very active such as climbing steps or running to catch a bus. In more
adventurous activities such as bungee jumping, gravitational potential energy is
converted into other types of energy. Even jumping from an aeroplane, the laws of
physics cannot be switched off.
In this chapter you will learn about work, energy and power, and use force–
displacement graphs to determine the amount of work done.
Content
INQUIRY QUESTION
How can the motion of objects be explained and analysed?
By the end of this chapter you will be able to:
• apply Newton’s first two laws of motion to a variety of everyday situations,
including both static and dynamic examples, and include the role played by
friction (friction = µFN) (ACSPH063) CCT
• investigate, describe and analyse the acceleration of a single object subjected to a
constant net force and relate the motion of the object to Newton’s second law of
motion through the use of: (ACSPH062, ACSPH063)
-- qualitative descriptions CCT
-- graphs and vectors ICT N
-- deriving relationships from graphical representations including F = m a and
relationships of uniformly accelerated motion ICT N
• apply the special case of conservation of mechanical energy to the quantitative
analysis of motion involving: ICT N
-- work done and change in the kinetic energy
of an object undergoing accelerated
rectilinear motion in one dimension (W = Fnet s)
-- changes in gravitational
potential energy of an object in a uniform field
(∆U = mg ∆h )
• conduct investigations over a range of mechanical processes to analyse qualitatively and quantitatively the concept of average power (P = ∆tE , P = F v ),
including but not limited to: ICT N
-- uniformly accelerated rectilinear motion
-- objects raised against the force of gravity
-- work done against air resistance, rolling resistance and friction
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
5.1 Forces and friction
The force of friction dominates our lives. Along with the force of gravity, it effects
almost every interaction we have with the physical universe. Friction is often
discussed in negative terms, such as a source of heat, inefficiency or energy loss, but
without it we would not be able to perform even simple tasks like eating, walking, or
holding objects in our hands.
FRICTION
FIGURE 5.1.1 This magnetic levitation train in
China rides 1 cm above the track, so the frictional
forces are negligible. The train is propelled by
a magnetic force to a cruising speed of about
430 km h−1.
Friction is a force that opposes movement. There are times in everyday life when
having friction is essential. Water, snow or ice on the road reduces friction between
the tyres and the road, which can create dangerous situations where the driver
cannot steer properly or stop. In the snowfields, drivers fit chains to the wheels of
their cars to increase friction. The chains break through the snow and ice, and the
car is able to grip the road. Similarly, friction is required in the car’s brakes when
the driver wants to slow down. Brake pads are designed to maximise the friction
between them and the brake drum or disc.
Conversely, there are many situations in which friction is a problem. Consider
the moving parts in the engine of a car. Friction can reduce an engine’s efficiency
and fuel economy, and cause parts such as bearings and pistons to wear out quickly.
Special oils and other lubricants are used to prevent moving metal surfaces from
touching. Similarly, the very high speeds achieved by mag-lev trains like the one
shown in Figure 5.1.1 are possible only because they use magnetic fields to levitate
above the track, which almost eliminates friction.
Static and kinetic friction
When two surfaces are in contact with one another, there are two different types of
friction to consider. For example, suppose you want to push your textbook along the
table. As you start to push the book, you find that the book does not move at first.
You then increase the force that you apply. Suddenly, at a certain critical value, the
book starts to move. The maximum frictional force
that must be overcome before
the book starts to slide is called static friction, Fs.
Once the book begins to slide, a much lower force than static force is needed
to keep the book moving
at a constant speed. This force is called kinetic friction
and is represented by Fk . The graph in Figure 5.1.2 shows how the force required
to move an object changes as static friction is overcome.
Applied force
maximum
static friction
Fs⃗
kinetic friction
Fk⃗
Time
Fs⃗
overcoming friction
Fk⃗
motion
constant velocity
FIGURE 5.1.2 To start an object moving over a surface, the static friction between the object and the
surface must be overcome. This requires a larger force than is needed to maintain a constant velocity
once the object is moving.
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MODULE 2 | DYNAMICS
The difference between static and kinetic friction can be understood when
you consider that even very smooth surfaces are jagged at the microscopic
level. When the book is resting on the table, the jagged points of its bottom
surface have settled into the valleys of the surface of the table, and this helps
to resist attempts to slide the book. Once the book is moving, the surfaces do
not have any time to settle into each other, so less force is required to keep the
book moving.
Even when surfaces are microscopically smooth, friction arises from the forces
of attraction between the atoms and molecules of the two different surfaces that
are in contact. When the surfaces are stationary, the particles of one surface bond
weakly to the particles on the other surface. Before one surface can move across the
other, these bonds must be broken. This extra effort adds to the static friction force.
However, these bonds cannot form again once there is relative motion between the
surface, so the kinetic friction is lower than static friction.
Coefficient of friction
The amount of friction between two surfaces depends on two factors:
• the nature of the surfaces
• the normal force FN , which pushes the surfaces together.
friction = µFN
PHYSICSFILE CCT
where
Calculating the coefficient
µ is the coefficient of friction
FN is the normal force (in N)
Friction is a vector. Its direction is always in the opposite direction to the
motion of a moving object or the force applied to a stationary object.
The coefficient of friction is represented by the symbol μ (Greek mu). It is a
dimensionless constant that depends on the nature of the surfaces and whether
the situation involves static friction (μs) or kinetic friction (μk). Table 5.1.1 lists
some typical values for the coefficient of friction between different combinations of
dry surfaces.
TABLE 5.1.1
Coefficients of static friction (μs) and kinetic friction (μk) for different surfaces.
Surfaces
μs
μk
rubber on dry bitumen
1.0
0.8
glass on glass
0.9
0.4
rubber on wet bitumen
0.8
0.5
steel on steel
0.7
0.6
smooth wood on wood
0.5
0.2
ice on ice
0.1
0.03
steel on ice
0.02
0.01
human joints
0.01
0.01
It is possible to calculate the coefficient
of friction experimentally by first
rearranging the formula for friction:
µ = friction
FN
If you push an object at a constant
speed across a surface, you can use
Newton’s second law to calculate the
force applied. If you know the weight
force of the object you can then derive
the coefficient of kinetic friction.
The same experiment can be conducted
to find the coefficient of static friction
by finding the maximum force that
can be applied to an object without
moving it.
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
145
Worked example 5.1.1
CALCULATING FRICTION
A person pushes a wooden box weighing 150 N across a wooden floor.
150 N
Calculate the force of friction acting on the box if the coefficient of kinetic friction
between the box and the floor is 0.2.
Thinking
Working
Recall the definition of friction.
The normal force will be equal and
opposite to the weight of the box.
Substitute in the values for this
situation to find the kinetic friction.
Solve the problem, giving an answer
with appropriate units and direction.
friction = µFN
FN = 150 N upwards
Fk = 0.2 × 150
Fk = 30 N in the opposite direction to
the motion of the box
Worked example: Try yourself 5.1.1
CALCULATING FRICTION
A person pushes a wardrobe weighing 100 N from one room to another.
Calculate the force of friction acting on the wardrobe if the coefficient of kinetic
friction between the wardrobe and the floor is 0.5.
FRICTION AND NEWTON'S FIRST LAW
Friction is so much a part of our everyday experience that it is sometimes hard to
recognise where it is acting. The ancient Greek philosophers had no understanding
of friction; they believed that it was simply the natural tendency of every object to
come to rest. Isaac Newton challenged this assumption with his first law of motion
and the idea of inertia. Newton proposed that the natural tendency of objects is to
continue to move. This means if an object is slowing down or coming to a stop, there
must be an unbalanced force at work. In many cases this unbalanced force is friction.
Since we are so accustomed to seeing objects affected by friction, it can be useful to
study low-friction situations to get an idea of how objects act in the absence of friction.
An example is an air hockey table such as the one shown in Figure 5.1.3. In this game,
jets of air blow up from the table to hold the puck slightly above the surface of the
table, creating a nearly frictionless environment. If you hit the puck hard enough, it will
bounce back and forth across the table many times before it comes to a stop.
FIGURE 5.1.3
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MODULE 2 | DYNAMICS
An air hockey table creates a nearly frictionless playing surface for the puck.
Worked example 5.1.2
DETERMINING THE EFFECT OF FRICTION
An ice skater is skating straight across a
smooth ice rink at 3 m s−1 with very little
friction between their skates and
the ice.
If their skates suddenly hit a rough
patch of ice that produces a large amount
of friction, how does this affect the motion
of the skater?
Thinking
Working
Recall Newton’s first law of motion and
the concept of inertia.
An object will maintain a constant
velocity unless an unbalanced external
force acts on it.
Identify the motion of the skater.
The skater is travelling in a straight
line at 3 m s−1.
Identify the effect of friction when the
skates hit the rough ice.
Friction will cause the skates to
decelerate.
Determine the effect on the skater’s
motion.
With the addition of friction, the skates
would suddenly slow down. This
might cause the skater to fall over,
since inertia would cause the upper
part of their body to continue at a
constant speed.
Worked example: Try yourself 5.1.2
DETERMINING THE EFFECT OF FRICTION
An air hockey puck slides in a straight line across an air hockey table with
negligible friction.
If the air stopped blowing so that the puck was in contact with the table, what
would happen to the motion of the puck?
FRICTION AND NEWTON’S SECOND LAW
You will recall from Chapter 4 that Newton’s second law of motion describes
the relationship between the force applied to an object, the mass of the object,
and its acceleration. Because friction always opposes the motion of an object, it
causes objects to decelerate rather than accelerate. However, Newton’s second law
still applies.
GO TO ➤ Section 4.2 on page 129
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
147
Worked example 5.1.3
PHYSICSFILE N
CALCULATING DECELERATION DUE TO FRICTION
Relationship between
coefficient of friction and
deceleration
A skater with a mass of 50 kg is skating south on smooth ice when she crosses a
rough patch of ice that produces a coefficient of kinetic friction of 0.2 between the
skates and the ice.
For an object sliding across a flat
surface, there is a predictable
relationship between the coefficient
of kinetic friction and the object’s
deceleration.
Friction is given by µFN, and the
normal force is equal and opposite
to
of the object, i.e.
the weight
FN = −Fg = − mg.
Therefore, for an object of mass
m sliding on a flat surface with a
coefficient of kineticfriction μk,
the force of friction Fk is given by
Fk = −µk mg . The negative sign indicates
that kinetic friction acts in the opposite
direction to the motion.
To calculate the deceleration (negative
acceleration) caused by this force,
we
can use Newton’s second law: F = ma
The
force is friction, so
deceleration
ma = −µk mg
Cancelling
themass from both sides
gives a = −µk g .
This means that the rate of deceleration
of an object sliding on a flat surface
can be found simply by multiplying
the acceleration due to gravity by the
coefficient of friction.
a Calculate the force of friction between the skates and the rough patch of ice.
Thinking
Recall the definition of friction.
Calculate the normal force, which is
equal and opposite to the weight of
the skater.
Substitute in the values for this
situation into the definition of friction
to find the kinetic friction.
Solve the problem, giving an answer
with appropriate units and direction.
Working
friction = µFN
Fg = mg
= 50 × 9.8
= 490 N downwards
∴ FN = 490 N upwards
Fk = 0.2 × 490
Fk = 98 N north
b Calculate the deceleration of the skater caused by friction.
Thinking
Recall Newton’s second law
Transpose the formula to make
acceleration the subject and solve
for a .
Determine the deceleration.
Working
F = ma
a=
=
F
m
98
50
−2
a = 1.96 m s north
Worked example: Try yourself 5.1.3
CALCULATING DECELERATION DUE TO FRICTION
An air hockey puck with a mass of 100 g slides across an air hockey table. The
coefficient of kinetic friction is 0.5.
a Calculate the force of friction between the puck and the table.
b Calculate the deceleration of the puck caused by friction.
WS
2.4
Friction affecting acceleration
In many situations the effect of friction is to reduce the acceleration of an object that
is being acted on by other forces.
Worked example 5.1.4
CALCULATING THE ACCELERATION OF A CONNECTED BODY WITH AND
WITHOUT FRICTION
A 1.5 kg trolley cart is connected by a cord to a 2.5 kg mass. The cord is placed over
a pulley and the mass is allowed to fall under the influence of gravity.
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MODULE 2 | DYNAMICS
1.5 kg
2.5 kg
a Assuming that there is no friction between the cart and the table, and the
pulley is frictionless, determine the acceleration of the cart.
Thinking
Working
Recognise that the cart and the falling
mass are connected, and determine a
sign convention for the motion.
As the mass falls, the cart will move to
the right. Therefore both downwards
movement of the mass and rightwards
movement of the cart will be
considered positive motion.
Write down the data that are given.
Apply the sign convention to vectors.
m1 = 2.5 kg
m2 = 1.5 kg
g = 9.8 m s−2 downwards
= +9.8 m s−2
Determine the forces acting on the
system.
The only force acting on the combined
system of the cart and mass is the
weight of the falling mass.
Fnet = Fg = mg
= 2.5 × 9.8
= +24.5 N
Calculate the total mass being
accelerated.
This net force has to accelerate both
the cart and the falling mass.
m1 + m2 = 2.5 + 1.5
= 4.0 kg
Use Newton’s second law to determine
the acceleration of the cart.
a=
Fnet
m
=
24.5
4.0
= +6.1
= 6.1 m s−2 to the right
b If a frictional force of 8.5 N acts against the cart, what is the acceleration now?
Thinking
Working
Write down the data that is given.
Apply the sign convention to vectors.
m1 = 2.5 kg
m2 = 1.5 kg
g = 9.8 m s−2 down
= +9.8 m s−2
friction = 8.5 N left
= −8.5 N
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
149
Determine the forces acting on
the system.
There are now two forces acting on
the combined system of the cart and
mass: the weight of the falling mass
and friction.
Fnet = Fg + friction
= 24.5 + (−8.5)
= 16.0 N
= 16.0 N in the positive direction
Use Newton’s second law to determine
acceleration.
a=
Fnet
m
= 16.0
4.0
= 4.0 m s−2 to the right
Worked example: Try yourself 5.1.4
CALCULATING THE ACCELERATION OF A CONNECTED BODY WITH AND
WITHOUT FRICTION
A 0.6 kg trolley cart is connected by a cord to a 1.5 kg mass. The cord is placed over
a pulley and allowed to fall under the influence of gravity.
a Assuming that there is no friction between the cart and the table, and the
pulley is frictionless, determine the acceleration of the cart.
b If a frictional force of 4.2 N acts against the cart, what is the acceleration now?
SKILLBUILDER
N
Identifying systematic error
In everyday language, ‘error’ can mean any kind of mistake
or misjudgement, However, in science ‘error’ has a special
meaning. It refers to a situation where an experimental
result does not match an expected result because of a
deficiency in the experimental set-up. For example, if you
were conducting an experiment to calculate the acceleration
of a falling object (where the expected value is 9.8 m s−2)
and your experiment gave a result of 7.1 m s−2, then your
experiment had an error of 2.7 m s−2 (the difference between
the expected value and the measured value).
Experiments can be affected by two types of error:
random error and systematic error. Random error is
caused by random variations in the conditions of the
experiment, and cannot be predicted. On the other
hand, systematic error is caused by a problem in the
experimental design or equipment that affects every
measurement in the same way and has a predictable
effect on the results.
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MODULE 2 | DYNAMICS
Wherever possible, scientists try to identify systematic
errors so that they can eliminate them and improve their
experiments. Common causes of systematic error are:
•
incorrect calibration of instruments
•
zero error (where the instrument does not read zero
when it is not measuring any value)
•
friction or air resistance
•
repeated errors in experimental method, e.g. repeatedly
failing to adjust for parallax error.
To identify systematic errors, scientists ask the following
questions about their experiments:
•
Is there a pattern of error in the experiment, i.e. are all
or most of the experimental results higher or lower than
expected?
•
Would any of the common causes of systematic error
affect this experiment?
•
If yes to the question above, would the effect of these
explain the pattern of error?
Worked example 5.1.5
IDENTIFYING SYSTEMATIC ERROR
The apparatus shown in the figure below was used as an experiment to
demonstrate Newton’s second law by placing different masses in the trolley and
using an accelerometer to measure the acceleration of the trolley.
mass
1.5 kg
2.5 kg
The data from the experiment are shown in the table below.
Accelerating
mass (kg)
Mass of trolley
(kg)
Mass in trolley
(kg)
Total mass of
system (kg)
Acceleration
(m s−2)
2.5
1.5
0.0
4.0
3.6
2.5
1.5
0.5
4.5
3.2
2.5
1.5
1.0
5.0
2.9
2.5
1.5
1.5
5.5
2.6
2.5
1.5
2.0
6.0
2.4
2.5
1.5
2.5
6.5
2.2
2.5
1.5
3.0
7.0
2.1
a Construct a graph of the data to show that Newton’s second law applies in this
situation.
Recall Newton’s second law
Construct a graph of the
acceleration versus the total
mass of the system.
Working
Fnet = ma
Newton’s second law experiment
4.0
3.5
3.0
Acceleration (ms–2)
Thinking
2.5
2.0
1.5
1.0
0.5
2.0
4.0
6.0
total mass (kg)
8.0
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
151
Inspect the shape of the
graph and compare it to a
prediction based on Newton’s
second law.
GO TO ➤ Section 1.5, page 28
Work out what you would
have to graph to get a
straight line. Recall how to
find a mathematical model
from a linear relationship
(Section 1.5).
The accelerating force (the weight of the
accelerating mass) is constant, so according
to Newton’s second law the total mass of the
system should be inversely proportional to its
acceleration. So the graph should have the
shape of a hyperbola.
This matches the shape of the graph observed.
Newton’s second law predicts that Fnet = ma .
1
a = Fnet m
Therefore, ↑ ↑ ↑
y= m x
Make a new table of the
manipulated data.
1/Total mass (kg−1)
Acceleration (m s−2)
0.25
3.6
0.22
3.2
0.20
2.9
0.18
2.6
0.17
2.4
0.15
2.2
0.14
2.1
Plot the graph of
manipulated data and
draw a line of best fit.
Newton’s second law experiment
Acceleration (ms–2)
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.05
Interpret the graph.
0.10 0.15 0.20 0.25
1/total mass (kg–1)
0.30
The graph is a straight line, which shows the
acceleration is inversely proportional to the
total mass.
This agrees with Newton’s second law, so the
law applies in this situation.
b Use the data to calculate the magnitude of the net force acting on the trolley
and determine if there is a systematic error in this experiment.
Thinking
Working
Calculate the equation of the line
of best fit.
The equation of the line of best fit was
calculated using a spreadsheet:
y = 14x
Find the equation relating a and m.
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MODULE 2 | DYNAMICS
The regression line has the equation
y = 14x, so the equation is a = 14
m
Relate this to Newton’s second
law.
According to Newton’s second law,
Fnet
.
a= m
∴ Fnet = 14 N
Compare the measured value to
the expected value.
The weight of the acceleration mass is
given by:
Fnet = Fg = mg
= 2.5 × 9.8
= 25 N (to two significant figures)
Since the measured value is only 15 N,
there is a systematic error of
25 − 14 = 11 N.
Identify the likely cause of the
systematic error.
The systematic error is probably caused by
friction in the wheels of the trolley, between
the wheels and the table, and between the
cord and the pulley.
Worked example: Try yourself 5.1.5
IDENTIFYING SYSTEMATIC ERROR
The apparatus shown in the figure below was used as an experiment to
demonstrate a Newton’s second law by placing different masses in the trolley and
using an accelerometer to measure the acceleration of the trolley.
mass
1.5 kg
5.0 kg
The data from the experiment are shown in the table below.
Accelerating
mass (kg)
Mass of trolley
(kg)
Mass in trolley
(kg)
Total mass of
system (kg)
Acceleration
(m s−2)
5.0
1.5
0.0
6.5
5.2
5.0
1.5
0.5
7.0
4.9
5.0
1.5
1.0
7.5
4.5
5.0
1.5
1.5
8.0
4.3
5.0
1.5
2.0
8.5
4.0
5.0
1.5
2.5
9.0
3.8
a Construct a graph of the data to show that Newton’s second law applies in
this situation.
b Use the data to calculate the net force acting on the trolley and determine
whether there is a systematic error in this experiment.
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
153
5.1 Review
SUMMARY
• Friction is a force that opposes movement.
• The amount of friction between two surfaces can
be calculated using: friction = µFN
• The frictional force between stationary surfaces is
called static friction. The frictional force between
sliding surfaces is called kinetic friction. Static
friction is greater than kinetic friction.
• Friction can cause moving objects to decelerate.
The amount of deceleration can be calculated
using Newton’s second law.
KEY QUESTIONS
1
2
3
In each of the following scenarios, identify whether it
would be better to increase or decrease the friction
between the surfaces:
a pushing a heavy wardrobe from one room to
another
b speed skating
c driving a car around a corner
d walking on a wet floor.
In each of the following scenarios, identify whether the
friction between the surfaces is static friction or kinetic
friction:
a a person skis down a slope
b a sailor gets friction burns when a rope slides
through his hands
c a car is parked on a hill.
A car is driving around a corner at 35 km h−1. If the
car drives over a puddle of oil that reduces the friction
between the car’s tyres and the road to almost zero,
how would this affect the car’s motion?
v⃗ = 35 km h–1
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MODULE 2 | DYNAMICS
4
Identify three different ways that systematic error can
be introduced to an investigation.
5
What horizontal force has to be applied to a wheelie
bin to wheel it to the street on a horizontal path
against a frictional force of 20 N at a constant speed of
1.5 m s−1?
6
A 0.50 kg metal block is attached by a piece of string
to a dynamics cart, as shown below. The block is
allowed to fall from rest, dragging the cart along.
The mass of the cart is 2.5 kg.
2.5 kg
0.50 kg
a If friction is ignored, what is the acceleration of the
block as it falls?
b If a frictional force of 4.3 N acts on the cart, what
is its acceleration?
5.2 Work
PHYSICS INQUIRY
CCT
Forces and motion
How can the motion of objects be explained and analysed?
COLLECT THIS …
•
CD
•
pop-top lid from disposable drink container
•
Blu Tack
•
balloon
DO THIS …
1 Place a bead of Blu Tack around the bottom of the pop-top lid.
2 Press this onto the CD, ensuring the centre hole is under the lid.
3 Blow up the balloon and twist the mouth to hold the air in.
4 Place the balloon over the closed pop-top lid. Let the balloon untwist.
5 Pop the lid open. The balloon will start deflating as the air is pushed out
through the centre of the CD.
6 Apply a force to the CD by pushing it for 5 cm. Measure how far the CD travels.
7 Repeat, adjusting the distance the CD is pushed to 10 cm, 20 cm and 40 cm.
RECORD THIS …
Describe the motion of the CD in the four sections.
Present your results in a table.
REFLECT ON THIS …
How can the motion of objects be explained and analysed?
How has the energy transformed through the different sections of motion?
The words ‘energy’ and ‘work’ are commonly used when describing everyday
situations. But these words have very specific meanings when used in a scientific
context, and they are two of the most important concepts in physics. They allow
physicists to explain phenomena on a range of scales, from collisions of subatomic
particles to the interactions of galaxies.
ENERGY
Energy is the capacity to cause change. A moving car has the capacity to cause a
change if it collides with something else. Similarly, a heavy weight lifted by a crane
has the capacity to cause a change if it is dropped. Energy is a scalar quantity; it has
magnitude but not direction.
The many different forms of energy can be broadly classified into two groups:
kinetic energy and potential energy.
Kinetic energy is energy associated with motion. Any moving object, like the
moving car in Figure 5.2.1, has kinetic energy.
FIGURE 5.2.1
A moving car has kinetic energy.
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
155
In some forms of kinetic energy, the moving objects are not easily visible. An
example of this is thermal energy, which is a type of kinetic energy related to the
movement of particles. Table 5.2.1 lists some different types of kinetic energy.
TABLE 5.2.1
Types of kinetic energy and their associated moving objects.
Type of kinetic energy
Type of motion
rotational kinetic
rotating objects
sound
air molecules
thermal
movement of atoms, ions or molecules due to temperature
translational kinetic
objects moving in a straight line
The potential energy of an object is associated with its position relative to
another object or within a field. For example, an object suspended by a crane has
gravitational potential energy because of its position in the Earth’s gravitational
field. Some examples of potential energy are listed in Table 5.2.2.
TABLE 5.2.2
Types of potential energy and their causes.
Type of potential energy
Cause
chemical
relative positions of atoms
elastic
attractive forces between atoms
gravitational
gravitational fields
magnetic
magnetic fields
nuclear
forces within the nucleus of an atom
The SI unit for energy
The SI unit for energy, the joule (J), is named after the English scientist James
Prescott Joule. He was the first person to show that kinetic energy could be converted
into heat energy. The energy represented by 1 J is approximately equivalent to the
energy needed to lift a 1 kg mass (e.g. 1 L of water) through a height of 0.1 m or
10 cm. More commonly, scientists work in units of kilojoules (1 kJ = 1000 J) or even
megajoules (1 MJ = 1 000 000 J).
WORK
When a force acts on an object and causes energy to be transferred or transformed,
work is being done on the object. For example, if a weightlifter applies a force to a
barbell to lift it, then work has been done on the barbell; chemical energy within the
weightlifter’s body has been transformed into the gravitational potential energy of
the barbell (Figure 5.2.2).
FIGURE 5.2.2 As a weightlifter lifts a barbell, chemical energy is transformed into gravitational
potential energy.
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MODULE 2 | DYNAMICS
Quantifying work
Work is a change in energy; that is, W = ∆E. More specifically, work is defined as
the product of the net force causing the energy change and the displacement of the
object in the direction of the force during the energy change.
Since work corresponds to a change in energy, the SI unit of work is also the
joule. The definition of work allows us to find a value for a joule in terms of other
SI units.
Since W = Fnet s , 1 J = 1 N × 1 m = 1 N m.
A joule is equal to a newton-metre. That is, a force of 1 N acting over a distance
of 1 m does 1 J of work.
Using the definition of a newton:
1 J = 1 N × 1 m = 1 kg m s−2 × 1 m = 1 kg m2 s−2
This defines a joule in terms of fundamental units.
Although both force and displacement are vectors, work is a scalar unit. So like
energy, work has no direction.
W = Fnet s
where
W is work (in J)
Fnet is net force acting on the
object (in N)
s is the displacement in the
direction of the force (in m).
PHYSICSFILE N
Units of energy
A number of non-SI units for energy are
still in use. When talking about the energy
content of food, sometimes a unit called
a Calorie (Cal) (Figure 5.2.3) is used. One
Calorie is defined as the amount of heat
required to increase the temperature
of 1 kg of water by 1°C, and is equal to
4200 J. Another unit called the calorie
(without a capital C) is one thousandth of
a Calorie, and is sometimes called ‘small
cal’. Neither of these units are used in
physics.
Electrical energy in the home is often
measured in kilowatt-hours (kW h). A
kilowatt-hour is a very large unit of energy:
1 kW h = 3 600 000 J or 3.6 MJ.
In atomic and nuclear physics, medical
physics and electronics, the electron volt
(eV) is often used. An electron volt is the
energy gained or lost when an electron
moves across a potential of one volt. It is
equal to about 1.6 × 10−19 J.
An old unit of energy that you might see
sometimes is the erg (from the Greek word
ergon for energy). An erg is a very small
unit of energy: 1 erg = 10−7 J. This unit is
no longer used in science.
FIGURE 5.2.3 The
amount of energy in a serving
of food is often measured in calories.
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Worked example 5.2.1
CALCULATING WORK
A person pushes a heavy box along the ground for 10 m with a net force of 30 N.
Calculate the amount of work done.
F⃗ = 30 N
10 m
Thinking
Recall the definition of work.
Working
W = Fnet s
Substitute in the values for this situation.
W = 30 × 10
Solve the problem, giving an answer with appropriate units
W = 300 J
Worked example: Try yourself 5.2.1
CALCULATING WORK
A person pushes a heavy wardrobe from one room to another by applying a
force of 50 N for a distance of 5 m. Calculate the amount of work done by the
person.
Work and friction
The energy change produced by work is not always obvious. Consider Worked
example 5.2.1, where 300 J of work was done on a box when it was pushed 10 m. A
number of energy outcomes are possible for this scenario.
• In an ideal situation, where there is no friction, all of this work would be
transformed into kinetic energy and the box would end up with a higher velocity
than before it was pushed.
• In most real situations, where there is friction between the box and the ground,
some of the work done would be converted into heat and sound because of the
friction, and the rest would become kinetic energy.
• In the limiting situation, where the force applied is exactly equal to the friction,
the box would slide at a constant speed. This means that its kinetic energy would
not change, so all of the work done would be converted into heat and sound.
A force with no work
FIGURE 5.2.4 According to the definition of
work, no work is done when a person holds a
box at a constant height.
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MODULE 2 | DYNAMICS
The mathematical definition of work has some unusual implications. One is that if
a force is applied to an object but the object does not move, then no work is done.
This seems counterintuitive; that is, it is not what you would expect. An example of
this is shown in Figure 5.2.4. While picking up a heavy box requires work, holding
the box at a constant height does no work on the box.
Assuming the box has a weight of 100 N and that it is lifted
from the ground to
a height of 1.2 m, the work done lifting it would be: W = Fnet s = 100 × 1.2 = 120 J.
In this case, energy is being transformed from chemical energy inside the person’s
body into the gravitational potential energy of the box.
However, when the box is held at a constant height, the net force is 0 N (because the
force applied by the lifter exactly balances gravity) and the displacement
is0 m (because
the box does not move). Therefore, the definition of work gives: W = Fnet s = 0 × 0 = 0 J.
So no work is being done on the box. Although there would be energy transformations
going on inside the person’s body to keep their muscles working, the energy of the box
does not change, so no work has been done on the box.
WORK AND DISPLACEMENT AT AN ANGLE
Sometimes, when a force is applied, the object does not move in the same direction as
the force. For example, in Figure 5.2.5, when a person pushes a pram, the direction of
the force is at an angle downwards, although the pram moves horizontally forwards.
F ⃗ cos θ
θ
S⃗
θ
F ⃗ sin θ
F⃗
FIGURE 5.2.5 When
a person pushes a pram, the force applied by the person can be resolved into a
horizontal force and a vertical force.
In this case, only the horizontal component of the push contributes to the work
being done on the pram. The vertical component of this force pushes the pram
downwards and is balanced by the normal reaction force from the ground.
In the situation of a person
pushing a pram, the person’s push
can be resolved into
a vertical component, F sin θ , and a horizontal component, F cosθ (Figure 5.2.5).
Substituting the horizontal component into the general definition for work gives:
W = F cosθ × s
= F s cos θ
W = F s cos θ
where θ is the angle
between
the force vector F and the
displacement vector s.
FORCE–DISPLACEMENT GRAPHS
As its name suggests, a force–displacement graph illustrates the way a force changes
with displacement. For a situation where the force is constant, this graph is simple.
For example, in Figure 5.2.6 the force–displacement graph for a person lifting a box
at constant speed is a flat horizontal line, showing that the force applied to the box
is constant throughout the lift.
20N
Force (N)
20N
25
20
15
10
5
0.2 0.4 0.6 0.8 1.0 1.2
Displacement (m)
FIGURE 5.2.6 The
force–displacement graph for this person lifting a box is a straight, horizontal line,
indicating that the force applied to the box is constant throughout the process.
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159
In contrast, the more you stretch a spring, the greater the force required to
keep stretching it. The force–displacement graph for a spring is also a straight line,
but this line shows a direct relationship between the force and the displacement
(Figure 5.2.7).
250
Force (N)
200
150
100
50
0.002
0.006
0.01
Displacement (m)
0.014
FIGURE 5.2.7 As a spring stretches, more force is required to keep stretching it. The force is
proportional to the extension.
WS
2.5
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MODULE 2 | DYNAMICS
WS
2.6
When a force changes with displacement, the amount of work done by the force
can be calculated from the area under its force–displacement graph.
For a constant force this is very simple. In the example of a person lifting a
box in Figure 5.2.6, the area can be found by counting the number of ‘force times
distance’ squares under the line. Similar strategies, either counting grid squares or
calculating the area of the shape under the graph, can also be used if the force varies
with the displacement in a spring.
5.2 Review
SUMMARY
• Energy is the capacity to cause a change.
• Work is done when energy is transferred or
transformed.
• Energy is conserved. It can be transferred or
transformed, but not created or destroyed.
• Work is done when a force causes an object to
be displaced.
• There are many different forms of energy.
These can be broadly classified as either
kinetic (associated with movement) or potential
(associated with the relative positions of
objects).
• Work is the productof net force and
displacement: W = Fnet s .
• If a force produces no displacement, no work is
done.
KEY QUESTIONS
1
Contrast the meanings of the words energy and
work.
2
Classify the type of energy that the following objects
possess as either kinetic or potential energy.
a the blades of a rotating fan
b a pile of bricks sitting in a wheel barrow
c hot water in a kettle
d the sound of music coming from a set of
headphones
e a car battery
3
When accelerating at the beginning of a ride, a cyclist
applies a force of 500 N for a distance of 20 m. What is
the work done by the cyclist on the bike?
4
In the case of a person pushing against a solid brick
wall, explain why no work is being done.
5
A cyclist does 2700 J of work when she rides her bike
at a constant speed for 150 m. Calculate the average
force the cyclist applies over this distance.
6
Two people push in opposite directions on a heavy box.
One person applies 50 N of force, the other applies 40 N
of force. There is 10 N of friction between the box and
the floor, which means that the box does not move. What
is the work done by the person applying 50 N of force?
7
If a weightlifter does 735 J of work against gravity when
lifting a 50 kg weight, how high off the ground would
the weight be lifted? (Remember: Fg = mg .)
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161
5.3 Energy changes
Any object that moves, such as those shown in Figure 5.3.1, has kinetic energy. Many
real-life energy interactions, such as throwing a ball, involve objects with kinetic
energy. Some of these, such as car collisions, have life-threatening implications. So
it is important to be able to quantify the kinetic energy of an object.
FIGURE 5.3.1
Any moving object, regardless of its size, has kinetic energy.
THE KINETIC ENERGY EQUATION
Kinetic energy is the energy of motion. It can be quantified by calculating the
amount of work needed to give an object its velocity.
Consider the dynamics
cart in Figure 5.3.2 of mass, m, starting at rest (i.e. u = 0).
It is pushed with force, F , which acts while the cart undergoes a displacement, s , and
gains a final velocity, v .
v⃗
u⃗ = 0
F⃗
F⃗
s⃗
The kinetic energy of a dynamics
cart can be calculated by considering the force (F )
acting on it over a given displacement ( s).
FIGURE 5.3.2
The work W done by the force causes a change in kinetic energy from its initial
value 12 mu 2 to a new value of 12 mv 2 .
The relationship between the work done and the change in kinetic energy can
be written mathematically as:
W = 12 mv 2 − 12 mu 2
where:
W is work (in J)
m is mass (in kg)
u is initial velocity (in m s−1)
v is final velocity (in m s−1).
This equation is known as the ‘work–energy theorem’ because it shows the
theoretical relationship between work and energy.
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MODULE 2 | DYNAMICS
In this situation, the cart was originally at rest (i.e. u = 0), so:
W = 12 mv 2
Assuming that no energy was lost as heat or noise and that all of the work is
converted into kinetic energy, this equation gives us a mathematical definition for
the kinetic energy of the cart in terms of its mass and velocity:
K = 12 mv 2
where K is kinetic energy (in J)
Because kinetic energy is a scalar quantity, the direction of the velocity is
unimportant, so it is possible to write the kinetic energy equation as:
K = 12 mv 2
SKILLBUILDER
N
Multiplying
vectors
where v is the speed.
Worked example 5.3.1
CALCULATING KINETIC ENERGY
When two vectors are multiplied
together (known as the dot
product of two vectors), the result
is a scalar variable. For instance,
in
the equation for work, W = Fnet s ,
the force vector is multiplied by the
displacement vector to produce
the scalar quantity of work.
A car with a mass of 1200 kg is travelling at 90 km h−1.
Calculate its kinetic energy at this speed.
Thinking
Working
−1
Convert the car’s speed to m s .
90kmh−1 =
90km
1h
=
90 000m
3600 s
= 25 m s−1
Recall the equation for kinetic energy.
K = 12 mv 2
Substitute the values for this situation
into the equation.
K=
State the answer with appropriate units.
K = 375 000 J
1
2
Similarly, in the equation for
kinetic energy, K = 12 mv 2, the two
velocity vectors are squared to
create the scalar value for energy.
× 1200 × 252
= 375 kJ
Worked example: Try yourself 5.3.1
CALCULATING KINETIC ENERGY
An 80 kg person is crossing the street, walking at 5.0 km h−1.
Calculate the person’s kinetic energy, giving your answer correct to two
significant figures.
APPLYING THE WORK–ENERGY THEOREM
The work–energy theorem can be seen as a definition for the change in kinetic
energy produced by a force:
W = 12 mv 2 − 12 mu 2 = K final − Kinitial = ∆K
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163
Worked example 5.3.2
CALCULATING KINETIC ENERGY CHANGES
A 2 tonne truck travelling at 100 km h−1 slows to 80 km h−1 before turning a corner.
a Calculate the work done by the brakes to make this change.
Thinking
Working
Convert the values into SI units.
000m
u = 100kmh−1 = 100km
= 100
1h
3600 s
= 28 m s−1
v = 80kmh−1 =
80km
1h
=
80 000m
3600 s
= 22 m s−1
m = 2 tonne = 2000 kg
Recall the work–energy theorem.
W = 12 mv 2 − 12 mu2
Substitute the values for this situation
into the equation.
W=
1
2
× 2000 × 222 − 12 × 2000 × 282
= 484 000 − 784 000
= −300 000
State the answer with appropriate
units.
The work done by the brakes was
−300 000 J, or −300 kJ.
Note: The negative value indicates that
the work has caused kinetic energy to
decrease.
b If it takes 50 m for this deceleration to take place, calculate the average force
applied by the truck’s brakes.
Thinking
Recall the definition of work.
Substitute the values for this situation
into the equation.
Transpose the equation to find the
answer.
Working
W = Fnet s
300 000 J = Fnet × 50 m
Note: The negative has been ignored
because work is a scalar.
J
Fnet = Ws = 3050000
= 6000 N
m
Worked example: Try yourself 5.3.2
CALCULATING KINETIC ENERGY CHANGES
As a bus with a mass of 10 tonnes approaches a school, it slows from 60 km h−1 to
40 km h−1.
a Use the work–energy theorem to calculate the work done by the brakes of the
bus. Give your answer to two significant figures.
b The bus travels 40 m as it decelerates. Calculate the average force applied by
the truck’s brakes.
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MODULE 2 | DYNAMICS
Notice that the definitions for kinetic energy and change in kinetic energy can
be derived entirely from known concepts: the definition of work, Newton’s second
law and the equations of motion. This might make kinetic energy seem a redundant
concept. However, using kinetic energy calculations can often make the analysis of
physical interactions quicker and easier, particularly in situations where acceleration
is not constant.
Worked example 5.3.3
CALCULATING SPEED FROM KINETIC ENERGY
A 1400 kg car, initially stationary, accelerates for 10 s. During this time its engine
does 900 kJ of work.
Assuming that all of the work is converted into kinetic energy, calculate the
speed of the car in km h−1 after 10 s. Give your answer correct to two significant
figures.
Thinking
Working
Assume that all of the engine’s work
has become kinetic energy. Recall the
equation for kinetic energy.
K = 12 mv 2
Transpose the equation to make v the
subject.
v=
2K
m
Substitute the values for this situation
into the equation.
v=
2 × 900 ×103
1400
State the answer in the required units.
v = 36 × 3.6 = 130 km h−1
= 36 m s−1
Δh ⃗
Worked example: Try yourself 5.3.3
CALCULATING SPEED FROM KINETIC ENERGY
A 300 kg motorbike has 150 kJ of kinetic energy.
Calculate the speed of the motorbike in km h−1. Give your answer correct to two
significant figures.
Fl⃗ = lift
DEFINING GRAVITATIONAL POTENTIAL ENERGY
One of the easiest forms of potential energy to study is gravitational potential
energy. Any object that is lifted above Earth’s surface has the capacity to cause
change due to its position in the Earth’s gravitational field. An understanding
of gravitational potential energy is essential to understanding common energy
transformations.
Gravitational potential is a measure of the amount of energy available to an
object due to its position in a gravitational field. The gravitational potential energy
of an object can be calculated from the amount of work that must be done against
gravity to get the object into its position.
Consider the weightlifter lifting a barbell in Figure 5.3.3. Assuming that the
barbell is lifted at a constant speed, then the weightlifter
must apply a lifting
force equal to the force due to gravity
on the barbell, Fg . The lifting force F l
is applied over a displacement ∆h , corresponding to the change in height of
the barbell.
Fg⃗ = weight
FIGURE 5.3.3 A weightlifter applies a constant
force over a fixed distance to give the barbell
gravitational potential energy.
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
165
The work done against gravity by the weightlifter is:
W = Fnet s = Fg ∆h
Because the force due to gravity is given by the equation Fg = mg , the work done
can be written as:
W = mg ∆h
The work carried out in this example has resulted in the transformation of
chemical energy within the weightlifter into gravitational potential energy. Therefore
the change in gravitational potential energy of the barbell is:
∆U = mg ∆h
The change in gravitational potential energy of an object, due to the work done
against a gravitational field, is given by:
∆U = mg∆h
where:
ΔU is the change in gravitational potential energy (in J)
m is the mass of the object (in kg)
g is the gravitational field strength (9.8 N kg−1 or 9.8 m s−2 downwards on Earth)
Δh is the change in height of the object (in m).
Worked example 5.3.4
CALCULATING GRAVITATIONAL POTENTIAL ENERGY
Because gravitational potential
energy is a scalar quantity, you
can remove the direction of
gravity and height in the equation
U = mgh.
A weightlifter lifts a barbell which has a total mass of 80 kg from the floor to a
height of 1.8 m above the ground.
Calculate the change in gravitational potential energy of the barbell during this
lift. Give your answer correct to two significant figures.
Thinking
Recall the formula for change in
gravitational potential energy.
Working
∆U = mg ∆h
Substitute the values for this situation
into the equation.
ΔU = 80 × 9.8 × 1.8
State the answer with appropriate
units and significant figures.
ΔU = 1411.2 J = 1.4 kJ
Worked example: Try yourself 5.3.4
CALCULATING GRAVITATIONAL POTENTIAL ENERGY
A grocery shelf-stacker lifts a 5 kg bag of dog food onto a shelf 30 cm above
the floor.
Calculate the gravitational potential energy of the bag when it is on the shelf.
Give your answer correct to two significant figures.
GRAVITATIONAL POTENTIAL ENERGY
AND REFERENCE LEVEL
When calculating gravitational potential energy, it is important to carefully define
the level that corresponds to U = 0. Often this can be taken to be the ground or sea
level, but the zero potential energy reference level is not always obvious.
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MODULE 2 | DYNAMICS
It does not really matter which point is taken as the zero potential energy
reference level, as long as the chosen point is used consistently throughout a
particular problem (Figure 5.3.4). If objects move below the reference level, then
their energies will become negative and should be interpreted accordingly.
The direction of the vector Δh is
from the object to the reference
level.
→
→
ΔU = mgΔh
= mg→ × (height above table)
→
ht
→
→
ΔU = mgΔh
→
= mg→ × (height hf
above floor)
By including the direction of
both gravity and height when
calculating the gravitational
potential energy, you can tell
whether the object moved above
or below the reference level.
FIGURE 5.3.4 In this situation, the zero potential energy reference point could be taken as either the
level of the table or the floor.
Worked example 5.3.5
CALCULATING GRAVITATIONAL POTENTIAL ENERGY RELATIVE TO A
REFERENCE LEVEL
A 60 kg weightlifter holds a 50 kg barbell 50 cm from the floor.
Calculate the increase in gravitational potential energy of the barbell after raising it
to 90 cm above the floor.
Use g = 9.8 N kg−1 and state your answer correct to three significant figures.
Thinking
Recall the formula for gravitational
potential energy.
Working
∆U = mg ∆h
Identify the relevant values for this
situation.
m = 50 kg
Only the mass of the bar is being lifted
(the weightlifter’s mass is a distractor).
Δh = 40 cm = 0.4 m
g = 9.8 N kg−1
Take the starting point of the barbell
as the zero potential energy level.
Substitute the values for this situation
into the equation.
∆U = 50 × 9.8 × 0.4
State the answer with appropriate
units and significant figures.
∆U = 196 J
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167
PHYSICSFILE CCT
Newton’s universal law of
gravitation
Gravitational field strength (N kg–1)
The equation ∆U = mg∆h is based
on the assumption that the Earth’s
gravitational field is constant.
However, Newton’s universal law of
gravitation predicts that the Earth’s
gravitational field decreases with
altitude (Figure 5.3.5). This decrease
only becomes significant far above the
Earth’s surface. Close to the surface the
assumption of a constant gravitational
field is valid.
12
Earth's gravitational field
strength with altitude
10
8
6
4
2
5 10 15 20 25 30 35
Altitude (× 1000 km)
FIGURE 5.3.5 The
Earth’s gravitational field
strength decreases with altitude.
Worked example: Try yourself 5.3.5
CALCULATING GRAVITATIONAL POTENTIAL ENERGY RELATIVE TO A
REFERENCE LEVEL
A father picks up his baby from its bed. The baby has a mass of 6.0 kg and the
mattress of the bed is 70 cm above the ground. When the father holds the baby
in his arms, it is 125 cm off the ground.
Calculate the increase in gravitational potential energy of the baby.
Use g = 9.8 N kg−1 and give your answer correct to two significant figures.
+ ADDITIONAL
Elastic potential energy
Another important form of potential energy is elastic potential energy. Elastic
potential energy can be stored in many ways; for example, when a spring is
stretched, a rubber ball is squeezed, air is compressed in a tyre, or a bungee
rope is extended during a jump.
Materials that have the ability to store elastic potential energy when work is
done on them, and then release this energy, are called elastic materials. Metal
springs and bouncing balls are common examples; however, many other
materials are at least partially elastic. If their shape is manipulated, items such
as our skin, metal hair clips and wooden rulers all have the ability to restore
themselves to their original shape once released.
Materials that do not return to their original shape and release their stored
potential energy are referred to as plastic materials. Plasticine is an example
of a very plastic material.
The elastic potential energy of an object, UE, is given by the formula:
where:
UE = 12 kx 2
k is a property of the elastic material called the spring constant
x is the amount of extension or compression of the material.
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MODULE 2 | DYNAMICS
5.3 Review
SUMMARY
• All moving objects have kinetic energy.
• Gravitational potential energy is the energy an
object has because of its position in a gravitational
field.
• The kinetic energy of an object is equal to the
work required to accelerate the object from rest to
its final velocity.
an object, ΔU,
• The gravitational potential energy of
is given by the equation ∆U = mg ∆h.
• The kinetic energy of an object is given by the
equation:
K = 12 mv 2
• Gravitational potential energy is calculated relative
to a zero potential energy reference level, usually
the ground or sea level.
• The work–energy theorem defines work as change
in kinetic energy:
W = 12 mv 2 − 12 mu2 = ∆K
• Because kinetic and potential energy are scalar
quantities, it is possible to calculate them without
using the direction of the velocity, acceleration or
height. i.e. K = 12 mv2 and U = mgh
KEY QUESTIONS
1
The mass of a motorbike and its rider is 230 kg. If they
are travelling at 80 km h−1, calculate their combined
kinetic energy.
2
A 1500 kg car is travelling at 17 m s−1. How much work
would its engine need to do to accelerate it to 28 m s−1?
3
A cyclist has a mass of 72 kg and is riding a bicycle
which has a mass of 9 kg. When riding at top speed,
their total kinetic energy is 5 kJ. Calculate the top
speed to two significant figures.
4
By how much is kinetic energy increased when the
mass of an object is doubled?
5
A 57 g tennis ball is thrown 8.2 m into the air. Use
g = 9.8 m s−2.
a Calculate the gravitational potential energy of the
ball at the top of its flight.
b Calculate the gravitational potential energy of the
ball when it has fallen halfway back to the ground.
6
When climbing Mount Everest (h = 8848 m), a
mountain climber stops to rest at North Base Camp
(h = 5150 m). If the climber has a mass of 65.0 kg,
how much gravitational potential energy will she gain
in the final section of her climb from the camp to
the summit? For simplicity, assume that g remains at
9.8 N kg−1.
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
169
5.4 Mechanical energy and power
FIGURE 5.4.1 A time-lapse photograph of a
bouncing tennis ball.
Mechanical energy is the energy that a body possesses because of its position or
motion. It is the sum of its kinetic energy and the potential energies available to it.
Gravitational potential energy, elastic potential energy and kinetic energy are all
forms of mechanical energy.
In many situations energy is transformed from kinetic energy to gravitational
potential energy, or vice versa. For example, when a tennis ball bounces, as shown
in Figure 5.4.1, much of its kinetic energy is converted into gravitational potential
energy and then back into kinetic energy again.
In situations where mechanical energy is conserved, this fact can be used to
analyse the position and motion of objects. If mechanical energy is not conserved,
this fact can be used to identify other important energy transformations that occur.
When considering energy changes, the rate at which work is done is often
important. For example, if two cars have the same mass, then the amount of work
required to accelerate each car from a standing start to 100 km h−1 will be the same.
However, the fact that one car can do this more quickly than another may be an
important consideration for some drivers when choosing which car to buy.
Physicists describe the rate at which work is done using the concept of power.
Like work and energy, this is a word that has a very specific meaning in physics.
MECHANICAL ENERGY
For a falling object, the mechanical energy is the sum of its kinetic and gravitational
potential energies:
E m = K + U = 12 mv 2 + mgh
This is a useful concept in situations where gravitational potential energy is
converted into kinetic energy or vice versa. For example, consider a tennis ball with
a mass of 60.0 g that falls from a height of 1.00 m. Initially its total mechanical
energy would consist of its kinetic energy, which would be 0 J, and the gravitational
potential energy at this height. Assuming g = 9.80 m s−2, then:
∆U = mg ∆h = 0.0600 × 9.80 × 1.00 = 0.588 J
In a system of bodies where
only mechanical energy (the
energy resulting from position and
motion) is considered, the total
mechanical energy of the system
is constant. That is,
Em = K + U = 0 + mgh
v = 0, so K = 0
(Em)initial = (Em)final
where Em is the mechanical
energy of the system.
This is called the law of
conservation of mechanical
energy.
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MODULE 2 | DYNAMICS
h=1m
Em = K + U =
FIGURE 5.4.2
1
mv2
2
+0
h = 0, so U = 0
A falling tennis ball provides an example of conservation of mechanical energy.
At the instant the ball hits the ground, the total mechanical energy would consist
of the gravitational potential energy available to it and its kinetic energy just prior
to hitting the ground. The gravitational potential energy is 0 J because the ball is
at ground level. To calculate kinetic energy, the ball’s velocity just before it hits the
ground can be calculated using one of the equations of motion:
v 2 = u 2 + 2a s
s = −1.00 m, a = −9.80 m s−2 and u = 0 m s−1, so:
2
v = 02 + 2(−9.80 × −1.00)
v = 19.6
= 4.43 m s−1 downwards
Therefore the kinetic energy of the tennis ball just before it hits the ground is:
K = 12 mv 2 =
1
2
× 0.06 × 4.432 = 0.588 J
Notice that the mechanical energy at the top of the 1.00 m fall is the same as the
mechanical energy at the bottom of the ball. At the top:
Em = K + U = 0 + 0.588 = 0.588 J
At the bottom:
Em = K + U = 0.588 + 0 = 0.588 J
In fact, mechanical energy is constant throughout the fall. Consider the tennis
ball when it has fallen halfway to the ground. At this point, h = 0.500 m and
v = 3.13 m s−1:
Em = K + U
= ( 12 × 0.0600 × 3.132 ) + (0.0600 × 9.80 × 0.500)
PHYSICSFILE N
Mechanical energy of a ball
falling through the air
In reality, as the tennis ball in
Figure 5.4.2 drops through the air,
a very small amount of its energy
is transformed into heat and sound,
and the ball will not quite reach a speed
of 4.43 m s−1 before it hits the ground.
This means that mechanical energy
is not entirely conserved. However,
this small effect can be considered
negligible for many falling objects.
= 0.294 + 0.294
= 0.588 J
Notice that, at this halfway point, the mechanical energy is evenly split between
kinetic energy (0.294 J) and gravitational potential energy (0.294 J).
Throughout the drop, mechanical energy has been conserved.
Worked example 5.4.1
MECHANICAL ENERGY OF A FALLING OBJECT
A basketball with a mass of 600 g is dropped from a height of 1.2 m.
Calculate the kinetic energy of the basketball at the instant it hits the ground.
Thinking
Working
Since the ball is dropped, its initial
kinetic energy is zero.
Kinitial = 0 J
Calculate the initial gravitational
potential energy of the ball.
Uinitial = mgh
= 0.600 × 9.8 × 1.2
= 7.1 J
Calculate the initial mechanical
energy.
(Em)initial = Kinitial + Uinitial
= 0 + 7.1
= 7.1 J
At the instant the ball hits the ground,
its gravitational potential energy is
zero.
Ufinal = 0 J
Mechanical energy is conserved in this
situation.
(Em)initial = (Em)final = Kfinal + Ufinal
7.1 = Kfinal + 0
Kfinal = 7.1 J
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
171
Worked example: Try yourself 5.4.1
MECHANICAL ENERGY OF A FALLING OBJECT
A 6.8 kg bowling ball is dropped from a height of 0.75 m.
Calculate the kinetic energy of the bowling ball at the instant it hits the ground.
USING MECHANICAL ENERGY TO CALCULATE VELOCITY
The speed of a falling object does not depend on its mass. This can be demonstrated
using mechanical energy.
Consider an object with mass m dropped from height h. At the moment
it is dropped, its initial kinetic energy is zero. At the moment it hits the ground,
its gravitational potential energy is zero. Using the law of conservation of
mechanical energy:
(Em)initial = (Em)final
Kinitial + Uinitial = Kfinal + Ufinal
0 + mgh = 12 mv 2 + 0
mgh = 12 mv 2
gh = 12 v 2
v 2 = 2 gh
v = 2 gh
The formula v = 2 gh can be used to find the velocity of a falling object as it hits
the ground. Note that the formula does not contain the mass of the falling object,
so if air resistance is negligible any object with any mass will have the same final
velocity when it is dropped from the same height.
Worked example 5.4.2
FINAL VELOCITY OF A FALLING OBJECT
A basketball with a mass of 600 g is dropped from a height of 1.2 m.
Calculate the speed of the basketball at the instant before it hits the ground.
Thinking
Working
Recall the formula for the velocity of a
falling object.
v = 2gh
Substitute the relevant values into the
formula and solve.
v = 2 × 9.8 × 1.2 = 4.8 m s−1
Interpret the answer.
The basketball is falling at 4.8 m s−1
just before it hits the ground.
Worked example: Try yourself 5.4.2
FINAL VELOCITY OF A FALLING OBJECT
A 6.8 kg bowling ball is dropped from a height of 0.75 m.
Calculate the speed of the bowling ball at the instant before it hits the ground.
172
MODULE 2 | DYNAMICS
USING CONSERVATION OF MECHANICAL ENERGY IN
COMPLEX SITUATIONS
The concept of mechanical energy allows physicists to determine outcomes in
situations where the equations of linear motion cannot be used. For example,
consider a pendulum with a bob of mass 400 g displaced from its mean position so
that its height has increased by 20 cm, as shown in Figure 5.4.3.
Because a falling pendulum converts gravitational potential energy into kinetic
energy, the conservation of mechanical energy applies to this situation. Therefore
the formula developed earlier for the velocity of a falling object can be used to find
the velocity of the pendulum bob at its lowest point.
v = 2 gh = 2 × 9.8 × 0.2 = 2 m s−1
The speed of the pendulum bob will be 2 m s−1 at its lowest point. However,
unlike the falling tennis ball, the direction of the bob’s motion will be horizontal
instead of vertical at its lowest point. The equations of motion relate to linear motion
and cannot be applied to this situation because the bob swings in a curved path.
Conservation of energy can also be used to analyse projectile motion; that is,
when an object is thrown or fired into the air with some initial velocity. Since energy
is not a vector, no vector analysis is required, even if the initial velocity is at an angle
to the ground.
0.2 m
0.4 kg
FIGURE 5.4.3 A falling pendulum provides an
example of conservation of mechanical energy.
Worked example 5.4.3
USING MECHANICAL ENERGY TO ANALYSE PROJECTILE MOTION
A cricket ball with a mass of 140 g is thrown upwards into the air from a height
of 1.5 m at a speed of 15 m s−1. Calculate the speed of the ball when it has
reached a height of 8.0 m.
Thinking
Working
Recall the formula for
mechanical energy.
Em = K + U = 12 mv 2 + mgh
Substitute in the values for the
ball as it is thrown.
(Em )initial = 12 (0.14 × 152 ) + (0.14 × 9.8 × 1.5)
Use the law of conservation of
mechanical energy to find an
equation for the final speed.
(Em )initial = (Em ) final
= 17.81 J
= 12 mv 2 + mgh
1 mv 2
2
= (Em )initial − mgh
v2 =
Solve the equation algebraically
to find the final speed.
v2 =
2 [( Em )initial − mgh]
m
2 (17.81 − 0.14 × 9.8 × 8.0)
0.14
= 97.63
v = 9.9ms−1
Interpret the answer.
The cricket ball will be moving at 9.9 m s−1
when it reaches a height of 8.0 m.
Worked example: Try yourself 5.4.3
USING MECHANICAL ENERGY TO ANALYSE PROJECTILE MOTION
An arrow with a mass of 35 g is fired into the air at 80 m s−1 from a height of 1.4 m.
Calculate the speed of the arrow when it has reached a height of 30 m.
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
173
PHYSICS IN ACTION
N
Energy transformations
The world record for the men’s pole vault is 6.16 metres
— about as high as a single-storey house. The women’s
record is 5.06 metres. During a pole vault, a number
of energy transformations take place. The vaulter has
kinetic energy as they run in. This kinetic energy is used
to bend the pole and carry them forwards over the bar.
As the pole bends, energy is stored as elastic potential
energy. The athlete uses this stored energy to increase
their gravitational potential energy and raise their centre
of mass over the bar. Once the pole has been released and
the bar has been cleared, the gravitational potential energy
of the athlete is transformed into kinetic energy as they
fall towards the mat. The energy changes are analysed by
making some assumptions about the athlete and the jump.
Consider a pole vaulter who has a mass of 60 kg and
runs in at 7.0 m s−1. Treat her as a point mass located at
her centre of mass, 1.2 m above the ground. She raises
her centre of mass to a height of 5.0 m as she clears the
bar, and her speed at this point is 1.0 m s−1.
As she plants the pole in the stop, the pole has not yet
been bent, so it has no elastic potential energy. Using
Em = K + U = 12 mv 2 + mgh, the vaulter’s total energy at this
point is 2180 J (Figure 5.4.4).
When the vaulter passes over the bar the pole is straight
again, so it has no elastic potential energy. Taking the
ground as zero height, and using the same relationship
as above, the vaulter’s total energy is now 2970 J. This
does not seem consistent with the conservation of energy
as there is an extra 790 J. The extra energy is from the
muscles in her body. Just before the athlete plants the
pole, she raises it over her head.
K
Ug
UE
ΣE
K
Ug
UE
ΣE
=
30
= 2940
=
0
= 2970 J
= 1470
= 710
=
0
= 2180 J
FIGURE 5.4.4 These diagrams, drawn at equal time intervals, indicate
that the vaulter slowed down as she neared the bar. Her initial kinetic
energy was stored as elastic potential energy in the bent pole, and was
finally transformed into gravitational potential energy and kinetic energy,
enabling her to clear the bar.
After the pole is planted but before she leaves the
ground, the athlete uses her arms to bend the pole
(Figure 5.4.5). She pulls downwards on the pole with one
arm while the other arm pushes upwards. The effect of
these forces is to do work on the pole and store some extra
elastic potential energy in it. This work will be converted
into gravitational potential energy later in the jump.
Energy has also been put into the system by the muscles
of the athlete as they do work after she has left the ground.
Throughout the jump, she uses her arm muscles to raise
her body higher.
At the end of the jump, she is actually ahead of the pole
and pushing herself up off it. In effect, she has been using
the pole to push up off the ground.
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MODULE 2 | DYNAMICS
FIGURE 5.4.5 As the pole is planted, the vaulter uses her arms to
bend the bar. The forces are shown by the vectors. By bending the
bar, the athlete has stored energy which will later be transformed into
gravitational potential energy.
Loss of mechanical energy
Mechanical energy is not conserved in every situation. For example, when a tennis
ball bounces a number of times, each bounce is lower than the one before it, as
shown in Figure 5.4.6.
While mechanical energy is largely conserved as the ball moves through the
air, a significant amount of kinetic energy is transformed into heat and sound
when the ball compresses and expands as it bounces. The ball will not have as
much kinetic energy when it leaves the ground as it did when it landed. This
means that the gravitational potential energy it can achieve on the second bounce
will be less than the gravitational potential energy it had initially, so the second
bounce is lower.
DEFINING POWER
FIGURE 5.4.6 Mechanical energy is lost with
each bounce of a tennis ball.
Power is a measure of the rate at which work is done. Mathematically:
P = Wt
Recall that when work is done, energy is transferred or transformed. So the
equation can also be written as:
P = ∆tE
where
P is the power (in W)
ΔE is the energy transferred or transformed (in J)
t is the time taken (in s).
For example, a person running up a set of stairs does exactly the same amount of
work as if they had walked up the stairs (i.e. W = ΔU = mgΔh). However, the rate of
energy change is faster for running up the stairs. Therefore, the runner is applying
more power than the walker (Figure 5.4.7).
PHYSICSFILE N
Efficiency of energy
transformations
staircase
running
staircase
walking
FIGURE 5.4.7 The runner and the walker do the same amount of work, but the power output of the
runner is higher than that of the walker.
Unit of power
The unit of power is named after the Scottish engineer James Watt, who is most
famous for inventing the steam engine. A watt (W) is defined as a rate of work of
one joule per second; in other words:
joule
1 watt = 11second
= 1 J s−1
In the real world, energy
transformations are never perfect –
there is always some energy ‘lost’.
The percentage of energy that is
effectively transformed by a device
is called the efficiency of that device.
A device operating at 45% efficiency
is converting 45% of its supplied
energy into the useful new form. The
other 55% is ‘lost’ or transferred to
the surroundings, usually as heat and
sound. This ‘lost’ energy is not truly
lost since energy cannot be created or
destroyed; rather, it becomes a form of
energy (e.g. heat and/or sound) that is
not useful.
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
175
Worked example 5.4.4
CALCULATING POWER
A box with a mass of 2 kg is carried up a 5 m staircase in 20 s.
F⃗g = 19.6 N
5m
Calculate the power required for this task. (Assume g = 9.8 m s−2.)
Thinking
Calculate the force applied.
Working
Fg = mg
= 2 × 9.8
Calculate the work done (i.e. the
energy transformed).
= 19.6 N
W = Fnet s
= 19.6 × 5
= 98 J = ΔE
Recall the formula for power.
P=
∆E
t
Substitute the appropriate values into
the formula.
P=
98
20
Solve.
P = 4.9 W
Worked example: Try yourself 5.4.4
CALCULATING POWER
PHYSICSFILE N
A weightlifter lifts a 50 kg barbell from the floor to a height of 2.0 m above the
ground in 1.4 s.
Horsepower
Calculate the power required for this lift. (Assume g = 9.8 m s−2.)
James Watt was a Scottish inventor and
engineer. He developed the concept
of horsepower as a way to compare
the output of steam engines with
that of horses, which were the other
major source of mechanical energy
available at the time. Although the
unit of one horsepower (1 hp) has
had various definitions over time, the
most commonly accepted value today
is around 750 W. This is actually a
significantly higher amount than an
average horse can sustain over an
extended period of time.
176
MODULE 2 | DYNAMICS
POWER AND UNIFORM ACCELERATION
Power is often of interest when comparing different machines because it gives a
measure of the rate at which energy transformations take place. Because power is
inversely proportional to the time taken for the energy transformation (i.e. P ∝ 1t ), a
more powerful machine does work more quickly than a less powerful one.
Worked example 5.4.5
UNIFORM ACCELERATION AND POWER
A car with a mass of 1200 kg accelerates uniformly from 0 to 60 km h−1 in 4.6 s.
What is the power being exerted by the engine of the car? Ignore rolling resistance
and air resistance, and state your answer correct to three significant figures.
Thinking
Working
−1
Convert the car’s final speed to m s .
60kmh−1 =
60km
1h
=
60 000m
3600 s
= 16.67 m s−1
Calculate the change in kinetic energy
of the car.
∆K = 12 mv 2 − 12 mu2
=
1
2
× 1200 × 16.672 −
1
2
× 1200 × 0
= 166.7 kJ
Substitute the appropriate values into
the power formula.
P=
=
Solve.
∆E
t
166.7 × 103
4.6
P = 36 kW
Worked example: Try yourself 5.4.5
UNIFORM ACCELERATION AND POWER
A car with a mass of 2080 kg accelerates uniformly from 0 to 100 km h−1 in 3.7 s.
What is the power exerted by the engine of the car? Ignore rolling resistance and
air resistance, and state your answer correct to three significant figures.
POWER, FORCE AND AVERAGE SPEED
In many everyday situations where friction is significant, a force is applied to an object
to keep it moving at a constant speed, e.g. pushing a wardrobe across a carpeted floor
or driving a car at a constant speed. In these situations, the power being applied can
be calculated directly from the force applied and the speed of the object.
Since P = ∆tE and W = Fnet s , then:
F s
P = nett = Fnet × st
s
Since
t is the definition of average velocity v , the power equation can be written
as P = F v.
Worked example 5.4.6
FORCE–VELOCITY FORMULATION OF POWER
A person pushes a heavy box along the ground at an average speed of 1.5 m s−1
by applying a force of 40 N.
v⃗ = 1.5 m s–1
F⃗ = 40 N
What amount of power does the person exert on the box?
PA
2.4
PA
2.4
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
177
Thinking
Working
Recall the force–velocity formulation of
the power equation.
P = Fv
Substitute the appropriate values into
the formula.
P = 40 × 1.5
Solve.
P = 60 W
Worked example: Try yourself 5.4.6
FORCE–VELOCITY FORMULATION OF POWER
A heavy wardrobe is pushed along the floor at an average speed of 2.2 m s−1.
The force of kinetic friction is 1200 N.
WS
2.7
What amount of power is required for this task? Give your answer correct to two
significant figures.
Air resistance and rolling resistance
In section 5.1 you learned about the difference between static friction and kinetic
friction. Both these types of friction occur when two surfaces are sliding or
about to slide across each other. However, other types of resistance to motion
can occur. For example, when an object rolls it experiences a retarding force
known as rolling resistance. Similarly, an object moving through air is
constantly colliding with air particles. This creates a retarding force known as
air resistance.
A car moving along a road experiences both air resistance and rolling
resistance. Rolling resistance comes from the interaction between the wheels
of the car and the road and from the various components such as cogs and
bearings that turn inside the car’s engine. Air resistance comes as the car
moves through the air in front of it. Modern cars have aerodynamic shapes
that allow the air to flow smoothly over the car, which reduces air resistance
(Figure 5.4.8).
FIGURE 5.4.8
Modern cars have aerodynamic shapes.
The factors that affect the amount of rolling resistance or air resistance
experienced by an object are complex. However, it is usually reasonable to assume
a constant value for these resistances at a particular speed.
178
MODULE 2 | DYNAMICS
Worked example 5.4.7
FORCE–VELOCITY FORMULATION OF POWER WITH AIR RESISTANCE/ROLLING
RESISTANCE
A car with a mass of 1500 kg is travelling at a constant speed of 20 m s−1. The
combination of air resistance and rolling resistance produces a retarding force
of 500 N.
Calculate the power required from the car’s engine to maintain this speed.
Thinking
Working
Remember that the forces on the car
must be in equilibrium if it is travelling
at a constant speed.
∑ F = Fforwards + Fdrag = 0
Fforwards = −Fdrag
Recall the force–velocity formulation of
the power equation.
P =Fv
Substitute the appropriate values into
the formula.
P = 500 × 20
Solve.
P = 10 000 W
Worked example: Try yourself 5.4.7
FORCE–VELOCITY FORMULATION OF POWER WITH AIR RESISTANCE/ROLLING
RESISTANCE
A car with a mass of 900 kg is travelling at a constant speed of 15 m s−1. Rolling
resistance and air resistance combine to oppose the motion of the car with a
force of 750 N.
What is the power output of the car’s engine at this speed?
+ ADDITIONAL
Drag
Air resistance is a special example of a force known as drag or fluid resistance.
This force slows the motion of any object that moves through a liquid or
gas. The amount of drag experienced by an object depends on a number of
factors, including the speed, the shape and cross-sectional area of the object,
and the density and viscosity of the fluid.
The following formula is used to calculate the drag force, FD:
FD = 12 ρv 2CD A
where:
ρ is the density of the fluid
v is the speed of the object relative to the fluid
A is the cross-sectional area of the object
CD is the drag coefficient
The drag coefficient CD is a dimensionless constant that is related to the
shape of the object and the way the fluid flows around it.
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
179
5.4 Review
SUMMARY
• Mechanical energy is the sum of the potential and
kinetic energies of an object.
• Power is a measure of the rate at which work is
• Considering kinetic energy and gravitational
• The power required to keep an object moving
at a constant speed can be calculated from the
product of the force applied and its average
speed: P = F v
• Rolling resistance is a force that opposes the
motion of rolling objects.
potential energy: Em = K + U =
1
2
done: P =
m v2 + m g h
• Mechanical energy is conserved in a falling object.
• Conservation of mechanical energy can be used to
predict outcomes in a range of situations involving
gravity and motion.
∆E
.
t
• Air resistance is a force that opposes the motion
of objects moving through the air.
• The final speed of an object falling from height h
can be found using the equation v = 2gh
• When a ball bounces, some mechanical energy is
transformed into heat and sound.
KEY QUESTIONS
1
A piano with a mass of 180 kg is pushed off the roof of
a five-storey apartment block, 15 m above the ground.
a Calculate the piano’s kinetic energy as it hits
the ground.
b Calculate the piano’s kinetic energy as it passes the
windows on the second floor, having fallen 10 m.
2
A branch falls from a tree and hits the ground with
a speed of 5.4 m s−1. From what height did the
branch fall?
3
A javelin with a mass of 800 g is thrown at an angle
of 40° to the horizontal. It is released at a height of
1.45 m with a speed of 28.5 m s−1.
a Calculate the javelin’s initial mechanical energy.
b Calculate the speed of the javelin as it hits
the ground.
180
MODULE 2 | DYNAMICS
4
A 1610 kg car accelerates from zero to 100 km h−1
in 5.50 s. Calculate its average power output over
this time.
5
A crane lifts a 500 kg concrete slab at a constant speed
of 5 m s−1. Calculate the power of the crane’s engine.
6
A 1700 kg car’s engine uses 40 kW of power to
maintain a constant speed of 80 km h−1. Calculate
the total force due to air resistance and rolling
resistance acting on the car.
Chapter review
KEY TERMS
air resistance
conservation of mechanical
energy
elastic
energy
friction
gravitational potential
energy
kinetic energy
kinetic friction
mechanical energy
pendulum
potential energy
power
rolling resistance
static friction
work
KEY QUESTIONS
1
A boy is using a horizontal rope to pull his billycart at a
constant velocity. A frictional force of 25 N also acts on
the billycart. What force must the boy apply to the rope?
A car is driving at 60 km h−1 towards an intersection.
The lights change to red, so the driver applies the
brakes and starts to decelerate. After slowing for a
few metres, the car reaches a section of road which is
covered with gravel; this reduces the amount of friction
between the car’s tyres and the road to almost zero.
Use Newton’s first law to describe the motion of the
car as it moves across the gravel.
The following information relates to questions 3–6.
While riding his bike, Lachie produces a forwards force of
150 N. The combined mass of Lachie and the bike is 100 kg.
2
3
If there is no friction or air resistance, what is the
magnitude of the acceleration of Lachie and the bike?
4
If friction opposes the bike’s motion with a force of
45.0 N, what is the magnitude of the acceleration of
the bike?
5
What must be the magnitude of the force of friction if
Lachie’s acceleration is 0.600 m s−2?
6
Lachie adds a 25.0 kg bag to his bike. What must
be the new forward force he produces in order to
accelerate at 0.800 m s−2 if friction opposes the motion
with a force of 30.0 N?
7
A car drives at a constant speed for 80 m. In order to
overcome friction, its engine applies a force of 2000 N.
Calculate the work done by the engine.
8
A crane lifts a 200 kg load from the ground to a height
of 30 m. What is the work done by the crane?
9
A person walks up a flight of 12 stairs. Each step is
240 mm long and 165 mm high. If the person has
a mass of 60 kg and g = 9.8 m s−2, what is the total
amount of work done against gravity?
10 If 4000 J is used to lift a 50.0 kg object vertically at a
constant velocity, what is the theoretical maximum
height to which the object can be raised?
11 A cricket ball with a mass of 156 g is bowled with a
speed of 150 km h−1. What is the kinetic energy of the
cricket ball?
12 If a 1200 kg car has 70 kJ of kinetic energy, what is
its speed?
13 The speed of an object is doubled. By how much does
its kinetic energy increase?
14 An 88 kg plumber digs a ditch 40 cm deep. By how
much does the plumber’s gravitational potential
energy change when he steps from the ground down
into the ditch?
15 How high does a brick with a mass of 2.0 kg need to
be lifted to have 100 J of gravitational potential energy?
16 A pencil is dropped from a table 76 cm above the
ground. What is the speed of the pencil as it hits the
ground?
17 A tennis ball is thrown straight upwards at 9.0 m s−1.
What is the maximum height it will reach?
18 A football with a mass of 0.43 kg is kicked off the
ground with a speed of 16 m s−1. Assuming there is no
air resistance, how fast will it be going when it hits the
crossbar, which is 2.44 m above the ground?
19 A bullet of mass 5 g strikes a ballistics pendulum of
mass 75 kg with speed v and becomes embedded in
the pendulum. When the pendulum swings back, its
height increases by 11 cm. For the following questions,
assume that the initial gravitational potential energy of
the pendulum was zero.
11 cm
a What was the gravitational potential energy of the
pendulum at the top of its swing?
b What was the kinetic energy of the pendulum when
the bullet first became embedded in it?
c What was the speed of the bullet just before it hit
the pendulum? (Assume that the bullet transferred
all of its kinetic energy to the pendulum.)
CHAPTER 5 | FORCES, ACCELERATION AND ENERGY
181
CHAPTER REVIEW CONTINUED
20 A crane can lift a load of 5 tonnes vertically through a
distance of 20 m in 5 s. What is the power of the crane?
21 A car with a mass of 720 kg (including driver)
accelerates from 0 to 100 km h−1 in 18.0 s. What is the
average power output of the car over this time?
22 If the engine of a 1400 kg car uses 25 kW to maintain
an average speed of 17 m s−1, calculate the force acting
on the car due to air and rolling resistance.
23 At the start of a 100 m race, a runner with a mass of
60 kg accelerates from a standing start to 8.0 m s−1 in a
distance of 20 m.
a Calculate the work done by the runner’s legs.
b Calculate the average force that the runner’s legs
apply over this distance.
24 When moving around on the Moon, astronauts find it
easier to use a series of small jumps rather than to
walk. If an astronaut with a mass of 120 kg (including
their space suit) jumps to a height of 10 cm on
the Moon, where the gravitational field strength is
1.6 m s−2, by roughly how much does his potential
energy increase?
25 A steel sphere is dropped from the roof of a five-storey
apartment block, 15 m above the ground.
a Calculate the sphere’s speed as it hits the ground.
b Calculate the sphere’s speed as it passes the
windows on the second floor, having fallen 10 m.
182
MODULE 2 | DYNAMICS
26 What power would be required to slide an object with
a mass of 5 kg at a constant speed of 3 m s−1 across
a rough surface with a coefficient of kinetic friction
of 0.5?
27 After completing the activity on page 155, reflect on
the inquiry question: How can the motion of objects be
explained and analysed?
In your response, discuss how you can predict
the motion of the CD hovercraft through energy
transformations, and what role is played by friction.
CHAPTER
Momentum, energy and
simple systems
Newton’s first law of motion describes the concept of inertia, which explains how a
mass resists changes to its motion unless an external force acts upon it.
A related concept is momentum, which defines the motion of an object in terms of
its mass and velocity. Objects with a larger momentum require a greater external
force to slow them down.
In this chapter you will see how the law of conservation of momentum can be used
to predict and explain motion. You will also look at the concept of impulse, which
relates force and the time over which it is applied to the change in momentum that
it produces.
Content
INQUIRY QUESTION
How is the motion of objects in a simple system dependent on
the interaction between the objects?
By the end of this chapter you will be able to:
• conduct an investigation to describe and analyse one-dimensional (collinear)
and two-dimensional interactions of objects in simple closed systems
(ACSPH064) CCT
• analyse qualitatively and predict, using the law of conservation of momentum
∑ mvbefore = ∑ mvbefore and, where appropriate, conservation of kinetic energy
2
2
, the results of interactions in elastic collisions
= ∑ 21 mv after
∑ 21 mv before
(
(
)
)
(ACSPH066) ICT N
• investigate the relationship and analyse information obtained from graphical
representations of force as a function of time
• evaluate the effects of forces involved in collisions and other interactions,
and analyse quantitatively the interactions using the concept of impulse
∆p = F ∆t ICT N
(
)
• analyse and compare the momentum and kinetic energy of elastic and inelastic
collisions (ACSPH066) ICT N
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
DS
2.1
PHYSICS INQUIRY
N
CCT
Momentum and
velocity
How is the motion of
objects in a simple
system dependent on
the interaction between
the objects?
6.1 Conservation of momentum
It is possible to understand some physics concepts without knowing the physics
terms or the equations that describe them. For example, you may know that once a
heavy object gets moving it is difficult to stop it, whereas a lighter object moving at
the same speed is easier to stop.
In Chapters 4 and 5 you saw how Newton’s laws of motion can be used to
explain these observations. In this section you will explore how these observations
can be related to the concept called momentum.
MOMENTUM
The momentum of an object relates to both its mass and its velocity. The elephant
running in Figure 6.1.1 has a large momentum because of its large mass, and the
faster it runs the more momentum it will have. An animal with a larger mass will
have more momentum than a smaller, lighter animal travelling at the same speed.
And the more momentum an object has, the more momentum it has to lose before
it stops.
COLLECT THIS…
•
medicine ball or other heavy
item
•
measuring tape
•
chalk
DO THIS…
1 Mark a line on the ground
with the chalk. This will be
the stopping line. Place the
measuring tape past the
stopping line to measure the
stopping distance.
2 Run fast towards the stopping
line.
3 Once you cross the line, stop as
quickly as possible. Record the
stopping distance.
4 Repeat at medium and slow
speeds.
5 Repeat again at fast, medium
and slow speeds, this time
carrying the heavy item.
RECORD THIS…
Describe how your stopping
distance changed with a change in
mass and velocity.
Present a table of your results.
REFLECT ON THIS…
How is the motion of objects in a
simple system dependent on the
interaction between the objects?
What were the variables of this
investigation?
184
MODULE 2 | DYNAMICS
FIGURE 6.1.1 Momentum is related to mass and velocity. The greater the mass or velocity, the harder
it is to stop or start moving.
The equation for momentum, p, is the product of the object’s mass m and its
velocity v.
p = mv
where:
p is momentum (in kg m s−1)
m is the mass of the object (in kg)
v is the velocity of the object (in m s−1)
The greater an object’s mass or velocity, the larger that object’s momentum will
be. Because velocity is a vector quantity, momentum is also a vector, so it must have
a magnitude, unit and direction. The direction of a momentum vector is the same
as the direction of the velocity vector. For calculations of change in momentum in a
single dimension, we can use the sign conventions of positive and negative.
The following derivation uses Newton’s second law to relate force to momentum.
Fnet = ma
=
=
=
The discovery of the neutron
(v − u)
m ∆t
mv − mu
∆t
∆p
∆t
This shows that a change in momentum is caused by the action ofa net force.
It also shows that the net force is equal to the change in momentum ∆p divided by
the time taken Δt for the change to occur, which is the rate of change of momentum.
Worked example 6.1.1
CALCULATING MOMENTUM
Calculate the momentum of a 60.0 kg student walking east at 3.5 m s−1.
Thinking
Working
Ensure that the variables are in their
standard units.
m = 60.0 kg
v = 3.50 m s−1 east
Apply the equation for momentum.
p = mv
= 60.0 × 3.50
= 210 kg m s−1 east
Worked example: Try yourself 6.1.1
CALCULATING MOMENTUM
Calculate the momentum of a 1230 kg car travelling north at 16.7 m s−1.
CONSERVATION OF MOMENTUM
The most important feature of momentum is that it is conserved in any interaction
between objects, such as a collision or an action−reaction. This means that the total
momentum in any system before the interaction will be equal to the total momentum
in the system after the collision. This is known as the law of conservation of
momentum and can be represented by the following relationship:
PHYSICSFILE N
In the early 1930s several physicists
observed that very high energy
radiation was emitted when beryllium
was bombarded with alpha particles
(Figure 6.1.2). They knew that this
radiation was not electrons, because it
did not carry an electric charge. They
assumed that it consisted of gamma
rays produced when the alpha rays
collided with beryllium nuclei.
But another physicist, James Chadwick,
thought that this radiation might be an
unknown particle with a mass similar
to the proton. The existence of this
particle had been predicted by Ernest
Rutherford, director of the Cavendish
Laboratory in England where Chadwick
was working. In 1932 Chadwick
conducted his own experiments to
measure the energy produced by
this radiation, and realised that the
conservation of momentum calculations
did not add up: gamma rays could
not produce the amount of energy
he observed.
He then bombarded boron with
alpha particles and used the law of
conservation of momentum to calculate
the mass of the unknown particle. He
found that it was 1.0067 times the
mass of a proton, which confirmed
that it was a new particle close in
mass to a proton but with no charge.
Other physicists quickly confirmed
his experiments, and the particle was
named the neutron.
beryllium atom
∑ pbefore = ∑ pafter
where ∑ p is the sum of the momentum of objects in a system.
+
+
+
To find the total momentum of objects in a system (whether before or after an
interaction) you can find the momentum of each object from its mass and velocity
and then sum the momentums of all the objects.
For collisions in one dimension, apply the sign convention of positive and
negative directions to the velocities and then use algebra to determine the answer
to the problem. For collisions in two dimensions you must resolve the momentum
of each object into perpendicular components, then sum the momentums in each
dimension, and finally combine the sums for each dimension into a single vector.
+
+
+
alpha particle
+
+
FIGURE 6.1.2 Experiments involving
bombarding beryllium with alpha particles
led to the discovery of the neutron.
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
185
Momentum in one-dimensional collisions
If two objects are colliding in one dimension, then the following equation applies:
∑ pbefore = ∑ pafter
m1u1 + m2u2 = m1v1 + m2 v 2
where:
m1 is the mass of object 1 (in kg)
u1 is the initial velocity of object 1 (in m s−1)
v1 is the final velocity of object 1 (in m s−1)
m2 is the mass of object 2 (in kg)
u2 is the initial velocity of object 2 (in m s−1)
v 2 is the final velocity of object 2 (in m s−1).
Worked example 6.1.2
CONSERVATION OF MOMENTUM
A 2.50 kg mass moving west at 4.50 m s−1 collides with a 1.50 kg mass moving
east at 3.00 m s−1.
Calculate the velocity of the 2.50 kg mass after the collision, if the 1.50 kg mass
rebounds at 5.00 m s−1 west.
Thinking
Working
Identify the variables using subscripts.
Ensure that the variables are in their
standard units.
m1 = 2.50 kg
u1 = 4.50 m s−1 west
v1 = ?
m2 = 1.50 kg
u2 = 3.00 m s−1 east
v2 = 5.00 m s−1 west
Apply the sign convention to the
variables.
m1 = 2.50 kg
u1 = −4.50 m s−1
v1 = ?
m2 = 1.50 kg
u2 = +3.00 m s−1
v2 = −5.00 m s−1
Apply the equation for conservation of
momentum involving two objects.
m1u1 + m2u2 = m1v1 + m2v2
(2.50 × −4.50) + (1.50 × 3.00) =
2.50 v1 + (1.50 × −5.00)
2.50v1 = −11.25 + 4.50 − (−7.50)
0.75
v1 = 2.50
= +0.30 m s−1
Apply the sign convention to describe
the direction of the final velocity.
186
MODULE 2 | DYNAMICS
v1 = 0.30 m s−1 east
Worked example: Try yourself 6.1.2
CONSERVATION OF MOMENTUM
A 1200 kg wrecking ball moving north at 2.50 m s−1 collides with a 1500 kg
wrecking ball moving south at 4.00 m s−1.
Calculate the velocity of the 1500 kg ball after the collision, if the 1200 kg ball
rebounds at 3.50 m s−1 south.
PHYSICS IN ACTION
N
Conservation of momentum in sports
The law of conservation of momentum has applications in many spheres of
human endeavour, including sports. This is most obvious in sports involving
collisions between objects, such as baseball, cricket, curling, lawn bowls, ten
pin bowling, bocce, pool, snooker, squash, table tennis and tennis.
FIGURE 6.1.3
Conservation of momentum is an important aspect of the sport of curling.
All of these sports require the athlete to cause one object to collide with
the other. The best players are able to control the momentum of the striking
object so that the magnitude and direction of momentum for both the
striking object and the target object after the collision result in a good score
or a positional advantage.
FIGURE 6.1.4 Controlling the momentum of this bowling ball will enable the player to pick up
a three-pin spare.
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
187
Momentum when masses combine
In the situations described so far, the two objects remain separate from each other.
However, it is possible for two objects to stick together when they collide. If two
objects combine when they collide, then the equation is modified to:
∑ pbefore = ∑ pafter
m1u1 + m2u2 = m1+ 2 v
where:
m1 is the mass of object 1 (in kg)
u1 is the initial velocity of object 1 (in m s−1)
m2 is the mass of object 2 (in kg)
u2 is the initial velocity of object 2 (in m s−1)
m1+2 is the combined mass of object m1 and m2 (in kg)
v is the final velocity of combined mass of m1 and m2 (in m s−1)
Worked example 6.1.3
CONSERVATION OF MOMENTUM WHEN MASSES COMBINE
A 5.00 kg lump of clay moving west at 2.00 m s−1 west collides with a 7.50 kg mass of
clay moving east at 3.00 m s−1. They collide to form a single, combined mass of clay.
Calculate the final velocity of the combined mass of clay.
Thinking
Working
Identify the variables using
subscripts and ensure that the
variables are in their standard units.
Add m1 and m2 to get m1+2.
m1 = 5.00 kg
u1 = 2.00 m s−1 west
m2 = 7.50 kg
u2 = 3.00 m s−1 east
m1+2 = 12.50 kg
v =?
Apply the sign convention to the
variables.
m1 = 5.00 kg
u1 = −2.00 m s−1
m2 = 7.50 kg
u2 = +3.00 m s−1
Apply the equation for conservation
of momentum.
m1+2 = 12.50 kg
v =?
∑ pbefore = ∑ pafter
m1u1 + m2u2 = m1+2v
(5.00 × −2.00) + (7.50 × 3.00) = 12.50 v
+ 22.50
v = −10.0
12.50
= +1.00 m s−1
Apply the sign convention to describe
the direction of the final velocity.
v = 1.00 m s−1 east
Worked example: Try yourself 6.1.3
CONSERVATION OF MOMENTUM WHEN MASSES COMBINE
A 90.0 kg rugby player running north at 1.50 m s−1 tackles an opponent with
a mass of 80.0 kg who is running south at 5.00 m s−1. The players are locked
together after the tackle.
Calculate the velocity of the two players immediately after the tackle.
188
MODULE 2 | DYNAMICS
Momentum in explosive collisions
It is also possible for one object to break apart into two objects in what is known
as an ‘explosive collision’. If one object breaks apart into two fragments when an
explosive collision occurs, then the equation becomes:
∑ pbefore = ∑ pafter
m1u1 = m2 v 2 + m3 v 3
where:
m1 is the mass of the object before breaking apart, object 1 (in kg)
u1 is the initial velocity of object 1 (in m s−1)
m2 is the mass of one fragment, object 2 (in kg)
v 2 is the final velocity of object 2 (in m s−1)
m3 is the mass of the other fragment, object 3 (in kg)
v 3 is the final velocity of object 3 (in m s−1)
PHYSICSFILE N
Conservation of momentum
in rockets and jets
If you hold a balloon filled with air, its
momentum is zero because it is at rest.
But if you release the balloon with its
neck open, air escapes, and this air has
momentum. The law of conservation
of momentum means that the balloon
must move in the opposite direction
with the same momentum. This will
continue until no more air is released
from the balloon.
0 = pballoon + pair
pballoon = − pair
Worked example 6.1.4
CONSERVATION OF MOMENTUM FOR EXPLOSIVE COLLISIONS
A 90.0 kg athlete throws a 1000 g javelin while running west at 7.75 m s−1.
Immediately after throwing the javelin her velocity is 7.25 m s−1 west.
Calculate the velocity of the javelin immediately after she throws it.
Thinking
Working
Identify the variables using subscripts
and ensure that the variables are in
their standard units. Note that m1 is
the sum of the bodies, i.e. the athlete
and the javelin.
m1 = 91 kg
u1 = 7.75 m s−1 west
m2 = 90 kg
v2 = 7.25 m s−1 west
m1 = 91 kg
u1 = −7.75 m s−1
m2 = 90 kg
v2 = −7.25 m s−1
Apply the equation for conservation
of momentum for explosive collisions.
m3 = 1.00 kg
v3 = ?
∑ pbefore = ∑ pafter
m1u1 = m2v2 + m3v 3
air
balloon
FIGURE 6.1.5 The momentum of the
air flowing out to the left is equal to the
forward momentum of the balloon to
the right.
m3 = 1.00 kg
v3 = ?
Apply the sign convention to the
variables.
∑ pbefore = ∑ pafter
91.0 × −7.75 = (90.0 × −7.25) + 1.00 v 3
− ( −652.5)
v 3 = −705.251.00
52.75
v 3 = −1.00
This is the principle upon which rockets
and jet engines are based. These
engines produce a high-velocity stream
of hot exhaust gases by burning a liquid
or solid fuel and forcing it through a
narrow opening. These gases have
a very large momentum as a result
of the high velocities involved, and
can accelerate rockets and jets to
high speeds as they acquire an equal
momentum in the opposite direction.
Rockets carry their own oxygen supply
for combustion, while jet engines use
the surrounding air supply.
= −52.8 m s−1
Apply the sign convention to describe
the direction of the final velocity.
v 3 = 52.8 m s−1 west
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
189
Worked example: Try yourself 6.1.4
CONSERVATION OF MOMENTUM FOR EXPLOSIVE COLLISIONS
PA
2.5
PA
2.5
A stationary 2000 kg cannon fires a 10.0 kg cannonball. After firing, the cannon
recoils north at 8.15 m s−1.
Calculate the velocity of the cannonball immediately after it is fired.
ELASTIC AND INELASTIC COLLISIONS
Momentum and total energy are always conserved in a closed system. Collisions
can either be elastic or inelastic. In elastic collisions, the total kinetic energy before
a collision is equal to the total kinetic energy after a collision. This is known as the
law of conservation of kinetic energy, and can be represented by the following
relationship.
2
2
= ∑ 12 mv after
∑ 12 mvbefore
2 1 2
1mu
2 1 1 + 2 m2u2
= 12 m1v12 + 12 m2v 22
where:
m1 is the mass of object 1 (in kg)
u1 is the initial velocity of object 1 (in m s−1)
v1 is the final velocity of object 1 (in m s−1)
m2 is the mass of object 2 (in kg)
u2 is the initial velocity of object 2 (in m s−1)
v 2 is the final velocity of object 2 (in m s−1)
In inelastic collisions, kinetic energy is converted into other types of energy
such as thermal and sound energy. If the collision is 100% inelastic, the final kinetic
energy will be a minimum and the objects will stick together. If the collision is 100%
elastic, then the final velocities will be a maximum.
If only the initial masses and velocities in a collision are known, you cannot
determine the energy lost during the collision. In this case you cannot tell whether
the collision was elastic or inelastic.
However, if both the initial and final masses and velocities of the systems are
known, you can determine how much energy was lost. In this case you can tell
whether the collision was elastic or inelastic.
190
MODULE 2 | DYNAMICS
Worked example 6.1.5
CONSERVATION OF KINETIC ENERGY
A 7.5 kg steel ball moving north at 2.2 m s−1 collided with a 0.5 kg rubber ball
moving south at 3.0 m s−1. Immediately after the collision, the steel ball had a
velocity of 1.55 m s−1 north and the rubber ball had a velocity of 6.75 m s−1 north.
Was this collision elastic, or was it inelastic?
Thinking
Working
Identify the variables using
subscripts. Ensure that the variables
are in their standard units.
m1 = 7.5 kg
u1 = 2.2 m s−1 north
v1 = 1.55 m s−1 north
m2 = 0.5 kg
u2 = 3.0 m s−1 south
v2 = 6.75 m s−1 north
Apply the sign convention to the
variables.
m1 = 7.5 kg
u1 = +2.2 m s−1
v1 = +1.55 m s−1
m2 = 0.5 kg
u2 = −3.0 m s−1
v2 = +6.75 m s−1
Apply the equation for conservation
of kinetic energy.
2
2
=∑ 12 mv after
∑ 12 mvbefore
2
= 12 m1u12 + 12 m2u22
∑ 12 mvbefore
=
1
2
× 7.5 × 2.22 + 12 × 0.5 × −3.02
= 18.15 + 2.25= 20.4 J
2
∑ 12 mvafter = 12 m1v12 + 12 m2v22
=
1
2
× 7.5 × 1.552 + 12 × 0.5 × 6.752
= 9.0 + 11.4
= 20.4 J
Determine whether the collision was
elastic or inelastic.
No energy was lost, so the collision was
elastic.
Worked example: Try yourself 6.1.5
CONSERVATION OF KINETIC ENERGY
An object with a mass of 2200 kg was travelling at 17.56 m s−1 north when it hit
a stationary object with a mass of 2150 kg. The two objects joined together and
moved away at 8.881 m s−1 north.
Was this collision elastic, or was it inelastic?
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
191
6.1 Review
SUMMARY
• Momentum is the product of an object’s mass and
velocity: p = mv
• The law of conservation of kinetic energy is
written as:
2
2
= ∑ 12 mv after
∑ 12 mv before
• Momentum is a vector quantity.
• Force is equal to the rate of change of momentum.
• In elastic collisions, no kinetic energy is lost.
• The law of conservation of momentum can be
applied to situations in which:
• In inelastic collisions, kinetic energy is lost.
-- two objects collide and remain separate:
∑ pbefore = ∑ pafter
m1u1 + m2u2 = m1v1 + m2v2
-- two objects collide and combine together:
∑ pbefore = ∑ pafter
m1u1 + m2u2 = m1+2v
-- one object breaks apart into two objects in an
explosive collision:
∑ pbefore = ∑ pafter
m1u1 = m2v2 + m3v 3
KEY QUESTIONS
1
Calculate the momentum of a 3.50 kg fish swimming
south at 2.50 m s−1.
2
Calculate the momentum of a 433 kg boat travelling
west at 22.2 m s−1.
3
Calculate the momentum of a 58.0 g tennis ball served
at 61.0 m s−1 towards the south.
4
Which object has the greater momentum: a medicine
ball of mass 4.5 kg travelling at 3.5 m s−1 or one of
mass 2.5 kg travelling at 6.8 m s−1?
5
A 70.0 kg man steps out of a stationary boat with
a velocity of 2.50 m s−1 forwards onto the nearby
riverbank. The boat has a mass of 400 kg and was
initially at rest. With what velocity does the boat begin
to move as the man steps out? Give the answer to
three significant figures.
192
MODULE 2 | DYNAMICS
6
A lawn bowls jack of mass 250 g is stationary on the
ground when it is hit by a 1.50 kg bowl travelling north
at 1.20 m s−1. If the jack’s velocity immediately after
the collision is 1.60 m s−1 north, with what velocity does
the bowl move immediately after hitting the jack? Give
the answer to two significant figures.
7
A railway wagon of mass 25 000 kg moving along a
horizontal track at 2.00 m s−1 runs into a stationary
locomotive and is coupled to it. After the collision, the
locomotive and wagon move off at a slow 0.300 m s−1.
What is the mass of the locomotive alone? Give the
answer to three significant figures.
8
A spacecraft of mass 10 000 kg, initially at rest, burns
5.0 kg of fuel−oxygen mixture in its rockets to produce
exhaust gases ejected at a velocity of 6000 m s−1.
Calculate the velocity that this exchange will give to
the spacecraft.
6.2 Change in momentum
In Section 6.1 the momentum of an object was defined in terms of its velocity and
mass. For each of the different collisions described in that section, the momentum
of the system was conserved. That is, when all of the objects involved in the collision
were considered, the total momentum before and after the collision was the same.
But for each separate object in those examples, momentum was not conserved
because they experienced a change in velocity because of the collision. When an
object changes its velocity, its momentum must also change, because momentum
is a vector quantity. An increase in velocity means an increase in momentum, and a
decrease in velocity means a decrease in momentum. You can also think of this as a
transfer of momentum from one object to another.
GO TO ➤ Section 6.1, page 184
CHANGE IN MOMENTUM IN ONE DIMENSION
It is easy to change the velocity of an object, and therefore its momentum. For
example, you can run faster or run slower; or you can press a little harder on the
pedals of a bike or press a little softer. You can also bounce an object off a surface;
the basketball in Figure 6.2.1 experiences a change in momentum when it changes
direction during a bounce. Change in momentum, ∆p, is also called impulse.
Consider an object moving in one dimension; that is, in a straight line. An impulse or
change in momentum in this situation can be calculated using the following equation.
Because momentum is a vector quantity, impulse or change in momentum is
also a vector, so it must have a magnitude, a unit and a direction.
Worked example 6.2.1
IMPULSE OR CHANGE IN MOMENTUM
A cyclist is riding a bike east at 8.20 m s−1 as they approach a stop sign. The total
mass of the cyclist and bike is 80.0 kg.
Calculate the impulse of the cyclist during the time it takes them to stop at the
stop sign.
Thinking
Working
Ensure that the variables are in
standard units.
m = 80 kg
u = 8.20 m s−1 east
v = 0 m s−1
Apply the sign convention to the
velocity vectors.
m = 80 kg
u = +8.20 m s−1
v = 0 m s−1
∆p = mv − mu
Apply the equation for impulse or
change in momentum.
impulse = ∆p
= pfinal − pinitial
= mv − mu
where:
∆p is the change in momentum
(in kg m s−1)
m is the mass (in kg)
v is the final velocity (in m s−1)
u is the initial velocity (in m s−1)
= (80 × 0) − (80 × 8.20)
= 0 − 656
= −656 kg m s−1
Apply the sign convention to describe
the direction of the impulse.
impulse = 656 kg m s−1 west
Worked example: Try yourself 6.2.1
FIGURE 6.2.1 A basketball undergoes a change
in momentum when it bounces, because the
direction of its velocity changes.
IMPULSE OR CHANGE IN MOMENTUM
A student with a mass of 55.0 kg hurries to class after lunch, moving at 4.55 m s−1
north. Suddenly she remembers that she has forgotten her laptop, and runs back
to her locker at 6.15 m s−1 south.
Calculate the impulse of the student during the time it takes her to turn around.
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
193
CHANGE IN MOMENTUM IN TWO DIMENSIONS
GO TO ➤ Section 3.1, page 100
The momentum of an object moving in two dimensions can change not only by
changing the magnitude of its velocity or by reversing the direction of motion,
but also by changing course in any direction. For example, the momentum of the
boat in Figure 6.2.2 changes because the boat changes direction. As you saw in
Chapter 3, a change in velocity in two dimensions can be calculated using geometry.
The equation for impulse can be manipulated slightly to illustrate where the change
in velocity is applied, as follows.
∆p = mv − mu
= m(v − u )
Worked example 6.2.2
FIGURE 6.2.2 Changing momentum in two
dimensions by changing direction.
CHANGE IN MOMENTUM IN TWO DIMENSIONS
A 65.0 kg kangaroo is moving at 3.50 m s−1 towards the west, then turns north
and moves at 2.00 m s−1.
Calculate the change in momentum of the kangaroo during this time.
Thinking
Working
Identify the formula for
calculating a change in
velocity ∆v .
∆v = final velocity − initial velocity
Draw the final velocity
vector v and the initial
velocity vector u separately.
Then draw the initial
velocity in the opposite
direction, which represents
the negative of the initial
velocity u.
Construct a vector diagram,
drawing v first and then
from its head draw the
opposite of u. The change
of velocity vector is drawn
from the tail of the final
velocity to the head of
the opposite of the initial
velocity.
Because the two vectors to
be added are at 90° to each
other, apply Pythagoras’
theorem to calculate the
magnitude of the change in
velocity.
Calculate the angle from the
north vector to the change
in velocity vector.
State the magnitude and
direction of the change in
velocity.
194
MODULE 2 | DYNAMICS
N
W
E
u⃗ = 3.50 m s–1 west
v⃗ = 2.00 m s–1 north
–u⃗ = 3.50 m s–1 east
S
–u⃗ = 3.50 m s–1 east
N
W
E
v⃗ = 2.00 m s–1 north
S
∆v 2 = 2.002 + 3.502
= 4.00 + 12.25
∆v = 16.25
= 4.03 m s−1
tanθ =
3.50
2.00
θ = tan−1 1.75
= 60.3°
∆v = 4.03 m s−1 N60.3°E
θ
∆v⃗
Identify the variables using
subscripts and ensure that
the variables are in their
standard units.
m = 65.0 kg
∆v = 4.03 m s−1 N60.3°E
Apply the equation for
impulse or change in
momentum.
∆p = mv − mu
= m(v − u )
= m∆v
= 65.0 × 4.03
= 262 kg m s−1
Apply the direction
convention to describe the
direction of the change in
momentum.
∆p = 262 kg m s−1 N60.3°E
Worked example: Try yourself 6.2.2
IMPULSE OR CHANGE IN MOMENTUM IN TWO DIMENSIONS
A 160 g pool ball rolling south at 0.250 m s−1 bounces off a cushion and rolls east
at 0.200 m s−1.
Calculate the impulse on the ball during its impact with the cushion.
WS
2.8
6.2 Review
SUMMARY
• The equation for impulse is: ∆p = mv − mu
• Change of momentum, ∆p, is also called impulse.
It is a vector quantity.
• Change in momentum in two dimensions can be
calculated using geometry.
• A change or transfer in momentum occurs when
an object changes its velocity.
KEY QUESTIONS
1
Calculate the impulse of a 9.50 kg dog that changes its
velocity from 2.50 m s−1 north to 6.25 m s−1 south.
2
Calculate the impulse of a 6050 kg truck as it changes
from moving west at 22.2 m s−1 to east at 16.7 m s−1.
3
The velocity of an 8.00 kg mass changes from
3.00 m s−1 east to 8.00 m s−1 east. Calculate the change
in momentum.
4
Calculate the change in momentum of a 250 g
grapefruit as it changes from rest to moving
downwards at 9.80 m s−1 after falling off a tree.
5
The momentum of a ball of mass 0.125 kg changes
by 0.075 kg m s−1 south. If its original velocity was
3.00 m s−1 north, what is the final velocity?
6
A 45.0 kg drone flying west at 45.0 m s−1 changes
course and flies north at 45.0 m s−1 at the same
altitude. Calculate the impulse of the drone during the
change in direction.
7
A marathon runner with a mass of 70.0 kg is running
with a velocity of 4.00 m s−1 north, and then turns a
corner to start running 3.60 m s−1 west. Calculate their
change in momentum.
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
195
6.3 Momentum and net force
GO TO ➤ Section 4.2, page 129
Chapter 4 on Newton’s second law of motion discussed the quantitative connection
between force, mass, time, and change in velocity. This relationship is explored
further in this section. The relationship
between change in momentum ∆p , the
period of time Δt, and net force Fnet helps to explain the effects of collisions and how
to minimise those effects. It is the key to providing safer environments, including in
sporting contexts such as that shown in Figure 6.3.1.
FIGURE 6.3.1
When two rugby players collide, they exert an equal and opposite force on each other.
Think about what it would feel like to fall onto a concrete floor. Even from one
metre high it would hurt. A fall from the same height onto a thick mattress would
not hurt at all. In both situations speed is the same, mass has not changed and
gravity provides the same acceleration. Yet the two would feel very different.
CHANGE IN MOMENTUM (IMPULSE)
PHYSICSFILE N
Momentum units
By using Newton’s second law the unit
newton second (N s) can be shown
to be equivalent to the unit for both
momentum and impulse (kg m s−1).
Given that 1 N = 1 kg m s−2
(from F = ma ), it follows that
1 N s = 1 kg m s−2 × s
so 1 N s = 1 kg m s−1
Therefore either N s or kg m s−1 can be
used as the unit for momentum.
196
MODULE 2 | DYNAMICS
According to Newton’s second law, a net force will cause a mass to accelerate. A
larger net force will create a faster change in velocity. The faster the change occurs
— that is, the smaller the period of time Δt when it occurs — the greater the net
force that produced that change. Landing on a concrete floor would change your
velocity very quickly. You would be brought to an abrupt stop within a very short
amount of time. But if you land on a thick foam mattress, the change occurs over
a much longer time. Therefore the force needed to produce the change is smaller.
Starting with
Newton’s second law, the relationship
between change in
momentum ∆p (impulse) and the variables of force Fnet (often written just as F ),
and time Δt becomes as follows.
F = ma
=m
( )
v −u
∆t
F ∆t = m( v − u )
= ∆p
where ∆p is the change in momentum (in kg m s−1)
These equations illustrate that, for a given change in momentum, the product of
force and time is constant. This relationship is the key to understanding the effects
of collisions.
Worked example 6.3.1
CALCULATING THE FORCE AND IMPULSE
A student drops a 160 g pool ball onto a concrete floor from a height of 2.00 m.
Just before it hits the floor, the velocity of the ball is 6.26 m s−1 downwards. Before
it bounces back upwards, there is an instant in time at which the ball’s velocity is
zero. The time it takes for the ball to change its velocity to zero is 5.0 milliseconds.
a Calculate the impulse of the pool ball, giving your answer to three
significant figures.
Thinking
Working
Ensure that the variables are in their
standard units.
m = 0.160 kg
u = 6.26 m s−1 down
v = 0 m s−1
Apply the sign and direction
convention for motion in one
dimension. Up is positive and down is
negative.
m = 0.160 kg
u = −6.26 m s−1
v = 0 m s−1
Apply the equation for change in
momentum.
∆ p = m(v − u )
= 0.160 × (0 − (−6.26))
= 1.0016 kg m s−1
Refer to the sign and direction
convention to determine the direction
of the change in momentum. This is
equal to the impulse.
impulse = ∆ p = 1.00 kg m s−1 upwards
b Calculate the average force that acts to cause the impulse.
Thinking
Use the answer to part a. Ensure
that the variables are in their
standard units.
Apply the equation for force.
Working
∆ p = 1.00 kg m s−1
Δt = 5.0 × 10−3 s
F ∆t = ∆ p
F = ∆∆pt
=
Refer to the sign and direction
convention to determine the direction
of the force.
1.00
5.0 × 10−3
= +200 N
F = 200 N upwards
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
197
Worked example: Try yourself 6.3.1
CALCULATING THE FORCE AND IMPULSE
A student drops a 56.0 g egg onto a table from a height of 60 cm. Just before it hits
the table, the velocity of the egg is 3.43 m s−1 down. The egg’s final velocity is zero,
which it reaches in 3.55 milliseconds.
a Calculate the impulse of the egg.
b Calculate the average force that acts to cause the impulse.
Worked example 6.3.2
CALCULATING THE FORCE AND IMPULSE (SOFT LANDING)
A student drops a 105 g metal ball onto a foam mattress from a height of 2.00 m.
Just before it hits the foam mattress, the velocity of the ball is 6.26 m s−1 down.
Before it bounces back up, there is an instant in time at which the ball’s velocity is
zero. The time it takes for the ball to change its velocity to zero is 0.360 seconds.
a Calculate the impulse of the pool ball.
Thinking
Working
Ensure that the variables are in their
standard units.
m = 0.105 kg
u = 6.26 m s−1 down
v = 0 m s−1
Apply the sign and direction convention
for motion in one dimension. Up is
positive and down is negative.
m = 0.105 kg
u = −6.26 m s−1
v = 0 m s−1
∆p = m(v − u )
Apply the equation for change in
momentum.
= 0.105 × (0 − (−6.26))
= 0.657 kg m s−1
Refer to the sign and direction
convention to determine the direction
of the change in momentum.This is
equal to the impulse.
impulse = ∆p = 0.657 kg m s−1 up
b Calculate the average force that acts to cause the impulse.
Thinking
Using the answer to part a, ensure that
the variables are in their standard units.
Apply the equation for force.
Working
∆p = 0.657 kg m s−1
Δt = 0.360 s
F ∆t = ∆p
F = ∆∆pt
=
Refer to the sign and direction convention
to determine the direction of the force.
198
MODULE 2 | DYNAMICS
0.657
0.360
= +1.83 N
F = 1.83 N up
Worked example: Try yourself 6.3.2
CALCULATING THE FORCE AND IMPULSE (SOFT LANDING)
A student drops a 56.0 g egg into a mound of flour from a height of 60 cm. Just
before it hits the mound of flour, the velocity of the egg is 3.43 m s−1 down. The egg
comes to rest in the flour in 0.325 seconds.
a Calculate the impulse of the egg.
b Calculate the average force that acts to cause the impulse.
•
•
•
•
From these worked examples you should notice a number of important things:
The change in momentum and the impulse were always the same.
Regardless of the surface that the object landed on, the impulse or change in
momentum remained the same.
The period of time was the main cause of the difference in the effects of the
different surfaces. Hard surfaces resulted in a short stopping time, and soft
surfaces resulted in a longer stopping time.
The effect of the period of time on the force was significant. A shorter time
meant a greater force, while a longer time meant a smaller force.
DETERMINING IMPULSE FROM A CHANGING FORCE
In the previous examples it was assumed that the force that acted to change the
impulse over a period of time was constant during that time. This is not always the
case in real situations. Often the force varies over the period of the impact, so there
needs to be a way to determine the impulse as the force varies.
An illustration of this is when a tennis player strikes a ball with a racquet. At the
instant the ball comes in contact with the racquet, the applied force is small. As the
strings distort and the ball compresses, the force increases until the ball has been
stopped. The force then decreases as the ball accelerates away from the racquet. A
graph of force against time is shown in Figure 6.3.2.
The
impulse affecting the ball during any time interval is the product of applied
force F and time Δt. The total impulse during the period of time the ball is in contact
with the racquet will be:
In a graph of force against time, the area under the curve is a function of the
height (force) and the width (time). So the total area under the curve in a force−
time graph is the total impulse for any collision, even when the force is not constant.
The concept of impulse is useful when dealing with forces during any collision,
because it links force and contact time; for example, when a person’s foot kicks a
soccer ball, or when a ball is hit by a bat or racquet. If the contact occurs over an
extended period of time as one or both objects deform, the average net force is used
because the forces change during the contact time.
The average net applied force can be found directly from the formula for
impulse. The instantaneous applied force at any particular time during the collision
must be read from a graph of force against time.
→
Impulse = Fav × ∆t
= area under graph
F (N)
impulse = Fav ∆t
where:
Fav is the average force applied during the collision (in N)
Δt is the total period of time the ball is in contact with the racquet (in s)
0
0.03
0.06
0.09
t (s)
0.12
0.15
FIGURE 6.3.2 The forces acting on the tennis
ball during its collision with the racquet are not
constant.
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
199
Worked example 6.3.3
CALCULATING THE TOTAL IMPULSE FROM A CHANGING FORCE
A student records the force acting on a rubber ball as it bounces off a hard
concrete floor over a period of time. The graph shows the forces acting on the ball
during its collision with the concrete floor.
80
Force (N)
70
60
50
40
30
20
10
0
1
2
3
4
5
6
7
8
Time (ms)
9 10 11 12 13
a Determine the force acting on the ball at a time of 9.0 milliseconds.
From the 9.0 millisecond
point on the x-axis go up to
the line of the graph, then
across to the y-axis.
Working
Force (N)
Thinking
80
70
60
50
40
30
20
10
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Time (ms)
The force is estimated by
reading the intercept of the
y-axis.
F = 48 N
b Calculate the total impulse of the ball over the 13 milliseconds of contact time.
Working
Break the area under the
graph into sections for which
you can calculate the area.
In this case, the graph can be broken into three
sections: A, B and C.
Force (N)
Thinking
80
70
60
50
40
30
20
10
B
A
C
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Time (ms)
200
MODULE 2 | DYNAMICS
area = A + B + C
Calculate the area of the
three sections A, B and C
using the equations for area
of a triangle and the area of a
rectangle.
= ( 12 bA × h) + (bB × h) + ( 12 bC × h)
=  12 × (2.0 × 10−3 ) × 60 + (6.0 × 10−3 ) × 60
+  12 × (5.0 × 10−3 ) × 60
= 0.060 + 0.36 + 0.15
= 0.57
The total impulse is equal to
the area.
impulse = area
Apply the sign and direction
convention for motion in one
dimension vertically.
impulse = 0.57 kg m s−1 upwards
= +0.57 kg m s−1
Worked example: Try yourself 6.3.3
CALCULATING THE TOTAL IMPULSE FROM A CHANGING FORCE
A student records the force acting on a tennis ball as it bounces off a hard concrete
floor over a period of time. The graph shows the forces acting on a ball during its
collision with the concrete floor.
80
70
Force (N)
60
50
40
30
20
10
0
1
2
3
4
5
6
7
Time (ms)
8
9
10 11 12 13
a Determine the force acting on the ball at a time of 4.0 milliseconds.
b Calculate the total impulse of the ball over the 13 millisecond contact time.
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
201
PHYSICS IN ACTION
ICT
Car safety
Vehicle safety is mainly about avoiding crashes. Research
shows that potential accidents are avoided 99% of the
time. This is mainly because of accident avoidance
systems such as antilock brakes. When a collision does
happen, passive safety features such as airbags and
crumple zones come into operation. Understanding
the theory behind accidents involves primarily an
understanding of impulse and force.
Airbags
The introduction of seatbelts allowed many more people
to survive car accidents. However, many survivors
still sustained serious injuries because of the rapid
deceleration in high-impact crashes, especially to the neck
because the head is not restrained. A further safety device
was needed to minimise these injuries.
Airbags are designed to inflate within a few milliseconds
of the start of a collision to reduce secondary injuries
during the collision. The airbag is designed to inflate only
when the car experiences an impact with a solid object at
18−20 km h−1 or more. This is so that minor bumps such
as parking accidents do not cause the airbags to inflate.
FIGURE 6.3.3 Airbags can prevent injuries by extending the period of
time a person take to stop.
The car’s computer control makes a decision within a few
milliseconds to detonate the gas cylinders that inflate the
airbags. The propellant in the cylinders inflates the airbags
while, according to Newton’s first law, the driver continues
to move towards the dashboard. As the passenger’s upper
body continues forwards into the airbag, the bag allows
the body to slow down over a longer time than would
otherwise be possible (Figure 6.3.3). This involves the
direct application of the concept of impulse. A comparison
of the forces applied to the occupant of a car with and
without airbags is shown in Figure 6.3.4.
(a)
Applied force (N)
(b)
no air bag
with air bag
0
202
20
40
60
MODULE 2 | DYNAMICS
80
Time (ms)
100
120
140
FIGURE 6.3.4 (a) The airbag extends the stopping time
and distributes the force required to decelerate the
mass of the driver or passenger over a larger area than
a seatbelt. (b) The force experienced by the occupant
of the car without an airbag is about double the force
experienced with an airbag.
6.3 Review
SUMMARY
• Forces can change during a collision.
• Newton’s second law describes the relationship
between impulse, force and the period of time:
impulse = ∆p = F ∆t
• The impulse over a period of time can be found
by calculating the area under the line on a force
versus time graph.
• The same mass changing its velocity by the
same amount will have a constant change in
momentum or impulse.
• The faster the velocity of a mass changes, the
greater the force required to change the velocity in
that period of time.
• The period of time during which an impulse
occurs is the cause of the difference in the forces
produced by different surfaces during a collision.
Harder surfaces result in a shorter stopping time,
and softer surfaces result in a longer stopping.
• The slower the velocity of a mass changes, the
smaller the force required to change the velocity
in that period of time.
• The effect of the period of time on the force is
dramatic. A shorter time means a greater force,
while a longer time means a much smaller force.
KEY QUESTIONS
A 45.0 kg mass changes its velocity from 2.45 m s−1
east to 12.5 m s−1 east in a period of 3.50 s.
a Calculate the change in momentum of the mass.
b Calculate the impulse of the mass.
c Calculate the force that causes the impulse of the
mass.
2
Using the concept of impulse, explain how airbags can
reduce injuries during a collision.
3
A student catches a 156 g cricket ball with ‘hard
hands’. Just before the student catches the ball,
its velocity is 12.2 m s−1 west. With hard hands, the
velocity of the ball drops to zero in just 0.100 s.
a Calculate the change in momentum of the cricket
ball.
b Calculate the impulse of the cricket ball.
c Calculate the average force on the cricket ball.
d The student now catches the same ball with
‘soft hands’. If the velocity of the ball drops from
12.2 m s−1 west to zero in 0.300 seconds, calculate
the average force on the cricket ball.
4
A stationary 130 g tee-ball is struck off the tee by a bat.
The ball and bat are in contact for 0.05 s, during which
time the ball is accelerated to a speed of 25 m s−1.
a What is the magnitude of the impulse the ball
experiences?
b What is the net average force acting on the ball
during the contact time?
c What is the net average force acting on the bat
during the contact time?
5
The following graph shows the net vertical force
generated as an athlete’s foot strikes an asphalt
running track.
a Estimate the maximum force acting on the athlete’s
foot during the contact time.
b Estimate the total impulse during the contact time.
1400
1200
1000
Force (N)
1
800
600
400
200
0
10
20
30
40
50
60
70
Time (ms)
6
A 25 g arrow buries its head 2.0 cm into a target on
striking it. The arrow was travelling at 50 m s−1 just
before impact.
a What change in momentum does the arrow
experience as it comes to rest?
b What is the impulse experienced by the arrow?
c What is the average force that acts on the arrow during
the period of deceleration after it hits the target?
7
Bicycle helmets are designed to reduce the force of
impact on the head during a collision.
a Explain how their design reduces the net force on
the head.
b Would a rigid ‘shell’ be as successful? Explain.
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
203
Chapter review
KEY TERMS
conserved
elastic collision
impulse
inelastic collision
law of conservation of
kinetic energy
law of conservation of
momentum
momentum
KEY QUESTIONS
Calculate the momentum of an elephant of mass
6000 kg running at velocity 40 km h−1.
2
A fly with a mass of 21 mg is travelling east at
1.8 m s−1. Calculate the momentum of the housefly.
3
A chestnut horse of mass 800 kg running at 20 m s−1
is passed by a grey horse of mass 600 kg running at
25 m s−1. Which horse has the greater momentum?
4
Calculate the change in momentum of a 155 kg mass
when its velocity changes from 6.50 m s−1 east to
3.25 m s−1 east in a period of 8.50 s.
5
Calculate the change in momentum of a 25.5 kg robot
when its velocity changes from 6.40 m s−1 forwards to
2.25 m s−1 backwards.
6
7
8
9
204
Julian is riding a skateboard north-east along a level
footpath, wearing a backpack. His mass is 40 kg, the
mass of the skateboard is 5 kg and the backpack is
2 kg. He is travelling at a constant speed of 5 m s−1
when he throws his backpack down so that it hits
the ground with zero horizontal velocity. What is his
velocity immediately after this happens?
An astronaut in a protective suit has a total mass of
154 kg. She throws a 40.0 kg toolbox away from the
space station. The astronaut and toolbox are initially
stationary. After being thrown, the toolbox moves at
2.15 m s−1. Calculate the velocity of the astronaut just
after throwing the toolbox.
A research rocket with a mass 250 kg is launched
vertically. It produces 50.0 kg of exhaust gases from
its fuel−oxygen mixture at a velocity of 180 m s−1 in a
2.00 s initial acceleration period.
a What is the velocity of the rocket after this initial
acceleration?
b What upwards force does this apply to the rocket?
c What is the net upwards acceleration acting on
the rocket? Assume that g = −9.81 m s−2 if required.
A bowling ball of mass 7 kg rolling north at 2 m s−1
collides with a pin of mass 0.5 kg. The pin flies north
at 5 m s−1. What is the velocity of the bowling ball after
the collision?
MODULE 2 | DYNAMICS
10 A 60 g tennis ball approaches a racquet with a velocity
of 50 m s−1. After being hit, the ball leaves the racquet
with a velocity of 30 m s−1 in the opposite direction.
Calculate the change in momentum of the ball during
the collision.
11 A 75.0 kg netball player moving west at 4.00 m s−1
changes velocity to 5.00 m s−1 north. Calculate her
change in momentum.
12 A 50 kg cyclist enters a roundabout travelling east at
8.5 m s−1. She leaves the roundabout heading south at
6.0 m s−1. Calculate her change in momentum between
entering and leaving the roundabout.
13 A billiard ball is travelling toward the edge of the
pool table at an angle of 30° from the edge. The ball
has a mass of 160 g and is travelling with a velocity
of 4.0 m s−1. The ball bounces off the edge with an
angle of 30° from the edge with a velocity of 3.0 m s−1.
Determine the change in momentum of the ball.
14 An athlete catches a 270 g volleyball by relaxing her
elbows and wrists and ‘giving’ with the ball. Just before
she catches the ball, its velocity is 5.60 m s−1 west. The
velocity of the ball then drops to zero in 1.00 second.
Calculate the average force exerted by the athlete on
the volleyball.
The following information relates to questions 15−17.
Jordy is playing softball and hits a ball with her softball
bat. The force versus time graph for this interaction is
shown below. The ball has a mass of 170 g.
500
Force (N)
1
0.015
Time (s)
0.040
15 Determine the magnitude of the change in momentum
of the ball.
16 Determine the magnitude of the change in momentum
of the bat.
17 Determine the magnitude of the change in velocity of
the ball.
18 A speed skater with a mass of 80 kg is travelling south
at 6.0 m s−1 when he collides with a second skater, who
has a mass of 70 kg and is travelling at 3.0 m s−1 in the
same direction. After the collision the velocity of the
first skater is 4.0 m s−1 south.
a How fast does the second skater leave the collision?
b Is this an elastic or inelastic collision?
20 A 50 g egg is dropped onto a hard floor. Immediately
before impact the egg is travelling at 5 m s−1. Using the
concept of momentum, describe a mechanism that
could prevent the egg from breaking. An eggshell can
withstand a force of up to 2 N.
21 After completing the activity on page 184, reflect on
the inquiry question: How is the motion of objects in a
simple system dependent on the interaction between
the objects?
In your response, discuss the concepts of momentum,
conservation of momentum and conservation of kinetic
energy.
19 A gardener is using a 5 kg mallet to hammer in a post.
The mallet is travelling downwards with a velocity of
5 m s−1 when it strikes the post. It rebounds with a
velocity of 1 m s−1 upwards.
a What is the impulse on the mallet?
b If the collision took 0.1 s, what is the magnitude and
direction of the force on the mallet?
CHAPTER 6 | MOMENTUM, ENERGY AND SIMPLE SYSTEMS
205
MODULE 2 • REVIEW
REVIEW QUESTIONS
Dynamics
WS
2.9
Multiple choice
1
3
Two boats are tied together as shown in the diagram
below. Boat A has twice the mass of boat B.
Boat A exerts a force of F on boat B in the positive
direction.
boat A
boat B
?
+F newtons
force on Moon
m
2m
Earth
What is the force on boat A from boat B?
A −0.5 F
B +F
C −F
D −2F
2
MR
2
The orbit of the Moon around Earth can be modelled as
circular motion in which the Moon can be considered to
orbit at a fixed height (radius) above the surface of Earth
and at a constant speed. Earth exerts a gravitational
force on the Moon that acts at right angles to the
velocity of the Moon as shown in the diagram below.
Moon
→
WS
2.10
A 60 kg student stands on a set of digital scales in an
elevator, as shown in the diagram below.
Which of the following statements about the Moon is
correct?
A The Moon experiences no change in kinetic energy
during an orbit.
B The Moon experiences no change in gravitational
potential energy during an orbit.
C Earth’s gravitational force does no work on the Moon.
D All of the above statements are correct.
4
scales
Two dynamics carts are being used in the physics
laboratory to study momentum. The two carts have the
same mass, m kg, and are travelling towards each other
with a speed v m s−1, as shown in the diagram.
v
v
m
a In which one or more of the following situations will
the digital scales show a reading of 60 kg?
A when the lift is travelling upwards at constant
velocity
B when the lift is travelling downwards at constant
velocity
C when the lift is stationary
D all of the above
b The lift travels upwards with an acceleration of a.
Taking upwards as the positive direction, which of
the following
is the correct expression for the normal
force (FN) on the student? m = mass of the student,
g = acceleration due to gravity.
A FN = mg + ma
B FN = mg − ma
C FN = ma − mg
D −FN = mg + ma
206
MODULE 2 | DYNAMICS
5
m
What is the magnitude of the total momentum of this
system?
A 0 kg m s−1
B mv kg m s−1
C 2mv kg m s−1
D −2mv kg m s−1
Hooke’s law says that the
extension x of a spring, can
be related to the force F used
toextend it by a spring
constant k by the formula F = k x . A single spring
has a spring constant of 10 N m−1. What mass must
be suspended from the spring to cause the spring to
stretch (extend) by 20 cm? Choose the closest answer.
A 2 g
B 20 g
C 200 g
D 2000 g
Which one or more of the following are examples of
contact forces?
A a tennis ball hitting the net
B a person pushing a wheelbarrow
C a person sitting in a chair
D a magnet moving iron filings
7
Which one or more of the following correctly describes
reaction pairs from Newton’s third law?
A Fracquet on ball = −Fball on racquet
B normal force = − weight force
C Fcup on table = −Ftable on cup
D gravitational force a chair exerts on Earth = −mg
8
9
Which one or more of the following are Newton’s laws of
motion?
A For every action there is an equal and opposite
reaction.
B An object will maintain a constant velocity unless an
unbalanced external force acts on it.
C The acceleration of an object is directly proportional
to the net force on the object.
D Constant velocity means the net force is not equal to
zero.
Which of the following correctly describes what happens
if an object maintains a constant velocity?
A The direction of the velocity does not change.
B The magnitude of the velocity does not change.
C The speed of the object does not change.
D The magnitude and direction of the velocity does not
change.
10 The following diagram shows three separate forces
acting on an object. What is the net force?
N
10 N
W
E
S
20 N
A
B
C
D
15 N
11 A car travels at a constant speed for 1 km. In order to
overcome friction, its engine applies a force of 1800 N.
How much work is done by the engine?
A 1.8 kJ
B 18 MJ
C 1.8 MJ
D 1.8 GJ
12 In which of the following situations is work not being
done?
A A diver jumps into a pool.
B A person rock climbs up a cliff face.
C A motorbike accelerates around a corner
D A person pushes against a solid wall.
13 Samuel carries a basket up to the second floor of his
house. The basket weighs 17.5 N and the flight of stairs
are 8.0 m long and inclined 35° to the horizontal. What
is the work done by Samuel against gravity?
A 80.3 J
B 114.7 J
C 140.0 J
D 46.8 J
14 A person picks up a box to a height of 1.4 m. Which
graph
(a) correctly describes this situation?
A
1.4
1.0
Force (N)
6
0.6
0.2
1.0
2.0
Displacement (m)
3.0
5 N west
15 N north
11.2 N N26.6°W
36.4 N N74.1°W
REVIEW QUESTIONS
207
MODULE 2 • REVIEW
B
(b)
15 A car with a mass of 1000 kg is travelling at 60 km h−1.
What is its kinetic energy?
A 13.9 kJ
B 180 kJ
C 139 kJ
D 1800 kJ
1.4
Force (N)
1.0
16 A tractor with a mass of 3000 kg is moving at 30 km h−1.
What is the magnitude of its momentum?
A 9000 kg m s−1
B 2500 kg m s−1
C 90 000 kg m s−1
D 25 000 kg m s−1
0.6
0.2
1.0
C
2.0
Displacement (m)
(c)
3.0
1.4
18 A 3.8 kg cat is running south at 2.8 m s−1 before changing
direction to 4.9 m s−1 north. What is the impulse of the
cat during this change in direction?
A 29.3 kg m s−1 south
B 29.3 kg m s−1 north
C 7.9 kg m s−1 south
D 7.9 kg m s−1 north
Force (N)
1.0
0.6
0.2
1.0
2.0
Displacement (m)
D
3.0
(d)
1.4
19 Two goats charge towards each other. One is 120 kg and
travelling at 2.1 m s−1 east, and the other is 100 kg and
travelling at 3.5 m s−1 west. They collide and their horns
lock them together. What is their velocity immediately
after the collision?
A 0.5 m s−1 west
B 0.5 m s−1 east
C 2.7 m s−1 west
D 2.7 m s−1 east
20 An 80 kg javelin thrower is running towards the throw
line at 7.6 m s−1 north. He throws the javelin and his
velocity immediately afterwards is 7.1 m s−1 north. If
the mass of the javelin is 1000 g, what is its velocity
immediately after being thrown?
A 7.1 m s−1 north
B 40.0 m s−1 north
C 47.6 m s−1 north
D 55.2 m s−1 north
Force (N)
1.0
0.6
0.2
Short answer
1.0
2.0
Displacement (m)
208
17 Two objects collide and stick together. Which equation
can be used to describe the collision?
A m1u1 + m2u2 = m1v1 + m2v2
B m1u1 = m2u2 + m3v 3
C m1u1 + m2u2 = m3v 3
D m1u1 = m2v2 + m3v 3
MODULE 2 | DYNAMICS
3.0
21 A 100 kg man is standing at rest on the ground. Use
g = −9.8 m s−2.
a Name the forces
acting on the man, using the
description FAB to describe the force by A on B.
b Indicate the relative magnitudes of these forces.
c Describe the reaction forces that form action−reaction
pairs with the forces on the man.
22 Three wooden blocks, each of weight 100 N, are stacked
one above the other on a table. Block C is on the table,
block B is in the middle and block A is on top. Use the
symbols T for table, E for Earth, and A, B and C for the
blocks to answer the following questions.
a Name the forces acting on block A and give the
magnitude and direction of each.
b Name the forces acting on block B and give the
magnitude and direction of each.
c Likewise, name the forces acting on block C and give
the magnitude and direction of each.
d A child quickly knocks the bottom block C out from
under the other two. Calculate the net force on each
of blocks A and B, and describe their motion.
23 A 100 kg trolley is being
pushed up a rough
30° incline
by a constant force F. The frictional force Ff between the
incline and the trolley is 110 N.
g⃗ = 9.8 m s–2
F⃗
30°
a Determine the magnitude of F that would move the
trolley up the incline at a constant velocity of 5.0 m s−1.
b Determine the magnitude of F that would accelerate
the trolley up the incline at 2.0 m s−2.
c Calculate
the acceleration of the trolley when
F = 1000 N.
25 An 800 N force is applied as shown to a 20.0 kg mass,
initially at rest on a horizontal surface. During its
subsequent motion the mass encounters a constant
frictional force of 100 N while moving along a horizontal
distance of 10 m.
F⃗ = 800 N
60°
g⃗ = 9.8 m s–2
26 The following diagram shows the trajectory of a 2.0 kg
sphere recorded by a physics student during a practical
investigation. The sphere is projected from a height of
2.0 m above the ground with initial speed v = 10 m s−1.
The maximum vertical height of the sphere is 5.0 m.
(Ignore friction and assume g = −9.8 N kg−1.)
C
B
A
2.0 m
5.0 m
3.0 m
a What is the total energy of the sphere just after it is
released at point A?
b What is the kinetic energy of the sphere at point B?
c What is the minimum speed of the sphere during its
flight?
d What is the total energy of the sphere at point C?
27 A child of mass 34 kg drops from a height of 3.50 m
above the surface of a trampoline. When the child lands
on the trampoline, it stretches so that she is 50 cm below
the initial level of the trampoline. (Use g = −9.8 m s−1).
a What is the spring constant of the trampoline?
Use W = 12 kx 2
FT⃗
10 kg
F⃗f = 100 N
a Determine the resultant horizontal force acting on the
20.0 kg mass.
b Calculate the work done by the frictional force.
c Calculate the work done by the resultant horizontal
force.
d Determine the change in kinetic energy of the mass.
e What is the final speed of the mass?
24 Two masses, 10 kg and 20 kg, are attached via a steel cable
to a frictionless pulley, as shown in the following diagram.
a→
20.0 kg
F⃗T
(
a→
20 kg
a Determine the acceleration for each mass.
b What is the magnitude of the tension in the cable?
)
b At what point does the child have maximum kinetic
energy?
c Describe the energy transformations that occur as the
child drops, bounces and rebounds.
28 A swimmer propels herself through the water with her
arms. Explain her motion in terms of Newton’s laws.
REVIEW QUESTIONS
209
MODULE 2 • REVIEW
30 An 86 kg football player travelling at 7.5 m s−1 collides
with a goalpost.
a Calculate the impulse of the post on the player.
b Is momentum conserved in this collision? Explain.
c The player hits his head on the goalpost during the
collision. Using Newton’s laws, explain why he might
suffer a concussion.
Extended response
31 The figure shows the velocity−time graph for a car of
mass 2000 kg. The engine of the car is providing a
constant driving force. During the 5.0 s interval the car
encounters a constant frictional force of 400 N.
The force−compression graph for the spring is shown
below. (In the following questions, ignore friction.)
v
spring
5.0 kg
12.0
10.0
Force (kN)
29 A spaceship with a mass of 20 tonnes (2.0 × 104 kg)
is launched from the surface of Earth, where g has a
−1
value of 9.8 N kg downwards. It lands on the Moon,
where g is 1.6 N kg−1 downwards. Assuming its mass
does not change during the voyage, what is the weight
of the spaceship when it is on Earth and when it is on
the Moon?
8.0
6.0
4.0
2.0
1.0 2.0 3.0 4.0 5.0 6.0
Compression (cm)
50
a Calculate the elastic potential energy stored in the
spring when it has been compressed by 2.0 cm.
b What is the elastic potential energy stored in the
spring when the trolley momentarily comes to rest?
c At what compression will the trolley come to rest?
d Explain why the trolley starts moving again.
e What property of a spring accounts for the situation
described above?
40
v (m s–1)
30
20
33 A nickel cube of mass 200 g is sliding across a horizontal
surface. One section of the surface is frictionless, while
the other is rough. The graph shows the kinetic energy K
of the cube versus distance, x, along the surface.
10
t (s)
4
6
a How much kinetic energy does the car have at
t = 5.0 s? Give your answer in megajoules.
b What is the net force acting on the car?
c What force is provided by the car’s engine during the
5.0 s interval?
d How much work is done on the car during the 5.0 s
interval?
e Determine the power output of the car’s engine
during the 5.0 s interval.
f How much heat energy is produced due to friction
during the 5.0 s interval?
32 A 5.0 kg trolley collides with a spring that is fixed
to a wall. During the collision the spring undergoes
a compression Δx, and the trolley is momentarily
brought to rest before bouncing back at 10 m s−1.
210
MODULE 2 | DYNAMICS
5.0
Kinetic energy (J)
2
4.0
3.0
2.0
1.0
1.0 2.0 3.0 4.0 5.0 6.0
x (cm)
a Which section of the surface is rough? Justify your
answer.
b Determine the speed of the cube during the first 2.0 cm.
c How much kinetic energy is lost by the cube between
x = 2.0 cm and x = 5.0 cm?
d What has happened to the kinetic energy that has
been lost by the cube?
e Calculate the value of the average frictional force
acting on the cube as it travels over the rough surface.
34 A naval gun with mass 1.08 × 105 kg fires projectiles
of mass 5.5 × 102 kg which leave the barrel at a speed
of 8.0 × 102 m s−1. The barrel of the gun is 20 m long,
and it can be assumed that the propellant acts on the
projectile for the time that it is in the barrel.
a Calculate the magnitude of the average acceleration
of the projectile down the barrel.
b Using Newton’s second law, calculate the average
force exerted by the propellant as the projectile
travels down the barrel.
c Calculate the momentum of the projectile as it leaves
the barrel.
d Calculate the recoil velocity of the gun.
e Calculate the average force of the propellant from the
change in momentum of the projectile.
f Calculate the average work done by the propellant
on the projectile, and compare this with the kinetic
energy gained by the projectile.
35 A goods train wagon of mass 4.0 × 104 kg travelling at
3.0 m s−1 in a shunting yard collides with a stationary
wagon of mass 1.5 × 104 kg, and the two wagons move
on coupled together at a reduced speed.
a Calculate the speed of the coupled wagons.
b Calculate the total momentum and kinetic energy of
the wagons before and after the collision.
c Is the collision elastic or inelastic? Explain your
answer.
d After the collision, the two wagons continue travelling
at a constant speed for 2 minutes. Calculate the
distance travelled.
e The wagons begin to decelerate at 1.5 m s−2. What is
the net force acting on the connected wagons?
REVIEW QUESTIONS
211
MODULE
Waves and thermodynamics
Wave motion involves the transfer of energy without the transfer of matter.
By exploring the behaviour of wave motion and examining the characteristics
of wavelength, frequency, period, velocity and amplitude, you can further your
understanding of the properties of waves.
This module also examines energy and its transfer, in the form of heat, from
one place to another. Thermodynamics is the study of the relationship between
energy, work, temperature and matter.
Outcomes
By the end of this module you will be able to:
• conduct investigations to collect valid and reliable primary and secondary data
and information PH11-3
• select and process appropriate qualitative and quantitative data and
information using a range of appropriate media PH11-4
• solve scientific problems using primary and secondary data, critical thinking
skills and scientific processes PH11-6
• communicate scientific understanding using suitable language and
terminology for a specific audience or purpose PH11-7
• explain and analyse waves and the transfer of energy by sound, light and
thermodynamic principles PH11-10
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
CHAPTER
Wave properties
Have you ever watched ocean waves heading toward the shore? For many people
their first thought when encountering a topic called ‘waves’ is to picture waves
moving across the surface of an ocean. These waves are created by a disturbance,
such as the action of wind on water or a boat moving through the water.
Waves are, in fact, everywhere. Sound, light, radio waves, waves in the string of an
instrument, the wave of a hand, the Mexican wave at a stadium, and the recently
discovered gravitational waves are all waves or wave-like phenomena.
Content
INQUIRY QUESTION
What are the properties of all waves and wave motion?
By the end of this chapter you will be able to:
• conduct a practical investigation involving the creation of mechanical waves in
a variety of situations in order to explain: CCT
-- the role of the medium in the propagation of mechanical waves
-- the transfer of energy involved in the propagation of mechanical waves
(ACSPH067, ACSPH070)
• conduct practical investigations to explain and analyse the differences
between: CCT
-- transverse and longitudinal waves (ACSPH068)
-- mechanical and electromagnetic waves (ACSPH070, ACSPH074)
• construct and/or interpret graphs of displacement as a function of time and as a
function of position of transverse and longitudinal waves, and relate the features of
those graphs to the following wave characteristics:
-- velocity
-- frequency
-- period
-- wavelength
-- wave number
-- displacement and amplitude (ACSPH069) ICT N
• solve problems and/or make predictions by modelling and applying the following
relationships to a variety of situations: ICT N
-- v = f λ
-- f = T1
-- k =
2π
λ
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
7.1 Mechanical waves
PHYSICS INQUIRY
N
CCT
Water waves
What are the properties of all waves and wave motion?
COLLECT THIS…
•
large glass container
•
open stand to hold up the container
•
water
•
white sheet of paper
•
strong light source
•
camera
•
ruler
•
tuning forks of different frequencies
•
graph paper
DO THIS…
1 Place the glass container on the stand. Fill it with 1 cm of water.
2 Shine a light from above and place the white paper underneath. Place the ruler on the paper running in the direction
of the longest side. Check that you can see the shadows of waves on the paper by dipping your finger in the water and
watching the paper. It might help to dim the lights in the room.
3 Hit the tuning fork with a rubber mallet. If there is no rubber mallet, hit it on something that is not hard, such as the
sole of your shoe.
4 Carefully put one prong in the water. Do not touch the container with the tuning fork.
5 As you put the tuning fork in the water, take a photo of the wave shadows on the paper.
6 Using the ruler to create a scale, measure the length of the waves.
7 Repeat with different frequency tuning forks, recording the frequency of each tuning fork.
RECORD THIS…
Describe the difference between the frequency and wavelength of a wave.
Present a graph of the wavelength against the inverse of the frequency.
REFLECT ON THIS…
What are the properties of all waves and wave motion?
If the frequency of the tuning fork was unknown, how could the frequency be measured?
What does the gradient represent on the graph of the wavelength vs the inverse of the frequency?
216
MODULE 3 | WAVES AND THERMODYNAMICS
Throw a stone into a pool or lake, and you will see circular waves form and move
outwards from the source as ripples, as shown in Figure 7.1.1. Stretch a cord out
on a table and wriggle one end back and forth across the table surface and another
type of wave can be observed. Sound waves, water waves and waves in strings are all
examples of mechanical waves. Mechanical waves, as opposed to electromagnetic
waves, cannot transmit energy through a vacuum. Mechanical waves are the focus
of this chapter.
MECHANICAL WAVES
Any wave that needs a medium (such as water) in which to travel is called a
mechanical wave. Mechanical waves include sound waves and water waves. They
can move over very large distances, but the particles of the medium have only very
limited movement.
If you watch a floating object such as a piece of driftwood, a surfboard or a boat
as a smooth wave goes past, you will see that the object moves up and down but does
not move forward with the wave. The movement of the object on the water reveals
how the particles in the water move as the wave passes: they move up and down
from an average position.
Mechanical waves transfer energy from one place to another through a medium.
The particles of the matter move up and down or backwards and forwards about an
average position, and this movement transfers the energy from one place to another.
For example, energy is given to an ocean wave by the action of the wind far out at
sea. The energy is transported by waves to the shore but (except in a tsunami) most
of the ocean water itself does not travel onto the shore.
FIGURE 7.1.1 The ripples in a pond indicate a
transfer of energy.
PHYSICSFILE N
Light waves
Light is also a type of wave known
as an electromagnetic wave. Unlike
mechanical waves, electromagnetic
waves do not require a medium. This
is why light from the Sun is able to
reach Earth through the vacuum
of space.
A wave involves the transfer of energy without the net transfer of matter.
PULSES VERSUS PERIODIC WAVES
A single wave pulse can be formed by giving a slinky spring or rope a single up and
down motion, as shown in Figure 7.1.2a. As the hand pulls upwards, the adjacent
parts of the slinky will also feel an upward force and begin to move upward. The
source of the wave energy is the movement of the hand.
If the up and down motion is repeated, each successive section of the slinky will
move up and down, moving the wave forward along the slinky as shown in Figure
7.1.2b. Connections between each loop of the slinky cause the wave to travel away
from the source, carrying with it the energy from the source.
(a)
one initial disturbance
wave pulse
(b)
continuous vibration at source
FIGURE 7.1.2 (a) A single wave pulse can be sent along a slinky by a single up and down motion.
(b) A continuous or periodic wave is created by a regular, repeated movement of the hand.
In a continuous wave like the one in Figure 7.1.2b, a continuous vibration of
the source causes the particles within the medium to oscillate about their average
position in a regular, repetitive (periodic) pattern. The source of a mechanical wave
is this repeated motion or vibration. The wave moves the energy from the vibration
through the medium.
CHAPTER 7 | WAVE PROPERTIES
217
Transverse waves
When waves travel in water or through a rope, spring or string, the particles within
the medium vibrate up and down in a direction perpendicular, or transverse, to
the direction of motion of the wave energy, as shown in Figure 7.1.3. For this reason
such a wave is called a transverse wave. When the particles are displaced upwards
from the average position (also called the resting position), they reach a maximum
positive displacement at a point called a crest. Particles below the average position
fall to a maximum negative position at a point called a trough.
wave travels right
wave source
original
water
level
PHYSICSFILE L
Water waves
Water waves are often classified
as transverse waves, but this is an
approximation. In practical situations,
transverse and longitudinal waves don’t
always occur in isolation. The breaking
of waves on a beach produces complex
wave forms which are a combination
of transverse and longitudinal waves
(Figure 7.1.5).
If you looked carefully at a cork bobbing
about in gentle water waves, you would
notice that it does not move straight up
and down, but it has a more elliptical
motion. It moves up and down, and
very slightly forwards and backwards
as each wave passes. However, since
this second aspect of the motion is
so subtle, in most circumstances it is
adequate to treat water waves as if they
were purely transverse waves.
cork now lower
crest
trough
FIGURE 7.1.3 As the continuous water wave moves to the right, the transverse up and down displacement
of the particles can be monitored using a cork. The cork simply moves up and down as the wave passes.
Longitudinal waves
In a longitudinal mechanical wave, the vibration of the particles within the medium
are in the same direction in which the wave is moving.You can demonstrate this type
of wave with a slinky by moving your hand backwards and forwards in a line parallel
to the length of the slinky, as shown in Figure 7.1.4.
rarefactions
compressions
Waves breaking on a beach
produce complex wave forms that are a
combination of transverse and longitudinal
waves.
FIGURE 7.1.5
WS
3.1
218
PA
3.1
PA
3.1
movement of hand
backwards and forwards
wave movement
particle movement
FIGURE 7.1.4 When the direction of the vibrations of the medium and the direction of travel of the
wave energy are parallel, a longitudinal wave is created. This can be demonstrated with a slinky.
As you move your hand, a series of compressed and expanded areas form along
the slinky. A compression is an area where the coils of the slinky come together.
Expansions are regions where the coils are spread apart. An area of expansion is
called a rarefaction. The compressions and rarefactions in a longitudinal wave
correspond to the crests and troughs of a transverse wave.
MODULE 3 | WAVES AND THERMODYNAMICS
7.1 Review
SUMMARY
• Vibrating objects transfer energy through waves,
travelling outwards from the source. Waves on
water, on a string and sound waves in air are
examples of mechanical waves.
• Mechanical waves can be either transverse or
longitudinal.
• In a transverse wave, the oscillations are
perpendicular to the direction in which the wave
energy is travelling. A wave in a string is an
example of a transverse wave.
• A wave may be a single pulse or it may be
continuous or periodic (successive crests and
troughs or compressions and rarefactions).
• In a longitudinal wave, the oscillations are parallel
to (along) the direction the wave energy is
travelling. Sound is an example of a longitudinal
wave.
• A wave only transfers energy from one point to
another. There is no net transfer of matter or
material.
KEY QUESTIONS
1
Describe the motion of particles in a medium as a
mechanical wave passes through the medium.
2
Which of the following statements are true and which
are false? For the false statements, rewrite them so
they become true.
a Longitudinal waves occur when particles of the
medium vibrate in the opposite direction to the
direction in which the wave is travelling.
b Transverse waves are created when the direction
of vibration of the particles is at right angles to the
direction in which the wave is travelling.
c A longitudinal wave is able to travel through air.
d The vibrating string of a guitar is an example of a
transverse wave.
3
The diagram below represents a slinky spring held at
point A by a student.
A
B
Draw an image of the pulse a short time after that
shown in the diagram and determine the motion
of point B. Is point B moving upwards, moving
downwards, or stationary?
4
The diagram below shows dots representing the
average displacement of air particles at one moment
in time as a sound wave travels to the right.
A B
compression
rarefaction
Describe how particles A and B have moved from their
equally spaced, undisturbed positions to form the
compression.
5
A sound wave is emitted from a speaker and heard by
Lee who is 50 m from the speaker. He made several
statements once he heard the sound. Which one or
more of the following statements made by Lee would
be correct? Explain your answer.
A Hearing a sound wave tells me that air particles
have travelled from the speaker to me.
B Air particles carried energy with them as they
travelled from the speaker to me.
C Energy has been transferred from the speaker to
me.
D Energy has been transferred from the speaker to
me by the oscillation of air particles.
CHAPTER 7 | WAVE PROPERTIES
219
WS
3.3
FIGURE 7.2.1 Waves can have different
wavelengths, amplitudes, frequencies, periods
and velocities, which can all be represented on
a graph.
7.2 Measuring mechanical waves
The features of a mechanical wave can be represented using a graph. In this
section you will explore how the displacement of particles within the wave can be
represented using graphs. From these graphs several key features of a wave can be
identified:
• amplitude
• wavelength
• frequency
• period
• speed.
Waves of different amplitudes and wavelengths can be seen in Figure 7.2.1.
DISPLACEMENT–DISTANCE GRAPHS
The displacement–distance graph in Figure 7.2.2 shows the displacement of all
particles along the length of a transverse wave at a particular point in time.
Particle
displacement
one wavelength, λ
A
λ
X
Y
P
A
B
Q
R
–A
Distance from source
FIGURE 7.2.2
amplitude, A
λ
A sine wave representing the particle displacements along a wave.
Look back at Figure 7.1.2b on page 217, showing a continuous wave in a slinky.
This ‘snapshot’ in time shows the particles as they move up and down sinusoidally
about a central rest position. As a wave passes a given point, the particle at that point
will go through a complete cycle before returning to its starting point. The wave has
the shape of a sine or cosine function, which you will recognise from mathematics.
A displacement–distance graph shows the position (displacement) of the particles
at any moment in time along the slinky about a central position.
From a displacement–distance graph, the amplitude and wavelength of a wave
are easily recognisable.
• The amplitude of a wave is the maximum displacement of a particle from the
average or rest position. That is, the amplitude is the distance from the middle
of a wave to the top of a crest or to the bottom of a trough. The total distance a
particle will move through in one cycle is twice the amplitude.
• Two particles on a wave are said to be in phase if they have the same
displacements from the average position and are moving in the same transverse
direction. The wavelength of a wave is the distance between any two successive
points in phase (e.g. points A and B, or X and Y, or P and R in Figure 7.2.2).
It is denoted by the Greek letter λ (lambda), and is measured in metres.
• The frequency, f, is the number of complete cycles that pass a given point per
second and is measured in hertz (Hz).
By drawing a series of displacement–distance graphs at various times, you can
see the motion of the wave. By comparing the changes in these graphs, the speed
and direction of the travelling wave can be found, as well as the direction of motion
of the vibrating particles.
• The wavenumber, k, is equivalent to the number of waves occurring over a
specified distance. It is calculated with the following formula.
220
MODULE 3 | WAVES AND THERMODYNAMICS
k = 2λπ
where: k is the wavenumber (in m−1)
λ is the wavelength of the wave (in m)
Worked example 7.2.1
DISPLACEMENT–DISTANCE GRAPH
The displacement–distance graph below shows a snapshot of a transverse wave
as it travels along a spring towards the right.
Particle displacement (cm)
20
Q
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
The sine curve is a mathematical
curve that describes a smooth
repetitive oscillation. It is
relevant to the study of sound,
AC electricity, simple harmonic
motion, electromagnetic waves,
and many other areas of physics.
The amplitude of the curve is
the distance from the midpoint
of the curve to the highest peak
or lowest trough, or half the
distance from the lowest to the
highest point.
−10
−20
N
GRAPHICAL
REPRESENTATION OF
A SINE CURVE
P
10
0
SKILLBUILDER
R
Distance (m)
Use the graph to determine the amplitude, wavelength and wavenumber of
this wave.
The wavelength of the curve is
the length of one complete wave.
Thinking
Working
Amplitude on a displacement–distance
graph is the distance from the average
position to a crest (P) or a trough (R).
Amplitude = 20 cm = 0.2 m
The period of the curve is the
time for one complete cycle of
the wave to occur.
Read the displacement of a crest or a
trough from the vertical axis. Convert
to SI units where necessary.
Wavelength is the distance for one
complete cycle. Any two consecutive
points that are in phase could be used.
0.8
The first cycle runs from the origin
through P, Q and R to intersect
the horizontal axis at 0.8 m. This
intersection is the wavelength.
Wavelength λ = 0.8 m
Wavenumber can be calculated using
k=
2π
λ .
wavelength
y
k=
2π
λ
=
6.3
0.8
= 7.9 m−1
0.4
amplitude
2
4
6
8
10
x
–0.4
–0.8
CHAPTER 7 | WAVE PROPERTIES
221
Worked example: Try yourself 7.2.1
DISPLACEMENT–DISTANCE GRAPH
Particle displacement (cm)
The displacement–distance graph below shows a snapshot of a transverse wave
as it travels along a spring towards the right.
2
1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
–1
–2
Distance (m)
Use the graph to determine the amplitude, wavelength and wavenumber of
this wave.
DISPLACEMENT–TIME GRAPHS
A displacement–time graph such as the one shown in Figure 7.2.3 tracks the position
of one point over time as the wave moves through that point.
Displacement
period
amplitude
Time
FIGURE 7.2.3 The graph of displacement versus time from the source of a transverse wave shows
the movement of a single point on a wave over time as the wave passes through that point.
The displacement–time graph looks very similar to a displacement–distance
graph of a transverse wave, so be careful to check the horizontal axis label.
Crests and troughs are shown the same way in both graphs. The amplitude is still
the maximum displacement from the average or rest position to either a crest or a
trough. But the distance between two successive points in phase in a displacement–
time graph represents the period of the wave, T, measured in seconds.
The period is the time it takes for any point on the wave to go through one
complete cycle (e.g. from crest to successive crest). The period of a wave is inversely
related to its frequency:
222
MODULE 3 | WAVES AND THERMODYNAMICS
T = 1f
where: T is the period of the wave (in s)
f is the frequency of the wave (in Hz)
The amplitude and period of a wave, and the direction of motion of a particular
particle, can be determined from a displacement–time graph.
Worked example 7.2.2
DISPLACEMENT–TIME GRAPHS
The displacement–time graph below shows the motion of a single part of a rope
(point P) as a wave passes by travelling to the right.
Displacement of P (m)
0.5
0
1
–0.5
2
3
4
Time (s)
Use the graph to find the amplitude, period, frequency and wavenumber of
the wave.
Thinking
Working
The amplitude on a displacement–time
graph is the displacement from the average
position to a crest or trough.
Maximum displacement is 0.5 m
Therefore amplitude = 0.5 m
Note the displacement of successive crests
and/or troughs on the wave and carefully
note the units on the vertical axis.
Period is the time it takes to complete
one cycle and can be identified on a
displacement–time graph as the time
between two successive points that are in
phase.
Period T = 2 s
Identify two points on the graph at the
same position in the wave cycle, e.g. the
origin and t = 2 s. Confirm by checking two
other points, e.g. two crests or two troughs.
Frequency can be calculated using:
f=
1
T
f=
1
T
=
1
2
= 0.5 Hz
CHAPTER 7 | WAVE PROPERTIES
223
Worked example: Try yourself 7.2.2
DISPLACEMENT–TIME GRAPHS
The displacement–time graph below shows the motion of a single part of a rope
as a wave passes travelling to the right. Use the graph to find the amplitude,
period and frequency of the wave.
Displacement of P (m)
0.1
0
0.25
0.5
–0.1
0.75
1.0
Time (s)
Use the graph to find the amplitude, period and frequency of the wave.
THE WAVE EQUATION
Although the speed of a wave can vary, there is a relationship between the speed of
a wave and other significant wave characteristics. Speed is given by:
v=
distance travelled
time taken
= ∆dt
This can be rewritten in terms of the distance of one wavelength, λ, in one
period, T, which will be:
v = fλ
where: v is the speed (in m s−1)
f is the frequency (in Hz)
λ is the wavelength (in m)
v = Tλ
and since
f = 1t
the relationship becomes: v = fλ
This is known as the wave equation and applies to both longitudinal and
transverse mechanical waves.
Worked example 7.2.3
THE WAVE EQUATION
What is the frequency of a longitudinal wave that has a wavelength of 2.0 m and
a speed of 340 m s−1?
Thinking
Working
The wave equation states that v = f λ.
Because you know both v and λ, the
frequency f can be found.
v = fλ
f = λv
Rewrite the wave equation in terms of f.
Substitute the known values and solve.
f=
=
v
λ
340
2.0
= 170 Hz
224
MODULE 3 | WAVES AND THERMODYNAMICS
Worked example: Try yourself 7.2.3
THE WAVE EQUATION
What is the frequency of a tranvserse wave that has a wavelength of
4.0 × 10−7 m and a speed of 3.0 × 108 m s−1?
Worked example 7.2.4
THE WAVE EQUATION
Calculate the period of a longitudinal wave that has a wavelength of 2.0 m and
a speed of 340 m s−1.
Thinking
Working
Rewrite the wave equation in
terms of T.
v = f λ and f =
Substitute the known values and solve.
v=
1
T
=
λ
T
T=
λ
V
T=
λ
V
=
1
T
×λ
2.0
340
= 5.9 × 10−3 s
Worked example: Try yourself 7.2.4
THE WAVE EQUATION
Calculate the period of a tranvserse wave that has a wavelength of
4.0 × 10−7 m and a speed of 3.0 × 108 m s−1.
WS
3.2
WS
3.3
PHYSICSFILE N
Seismic waves
On 28 December 1989 an earthquake devastated the region in
and around Newcastle, New South Wales. The earthquake was
rated at 5.6 on the Richter scale, with the epicentre at Boolaroo,
approximately 20 km west of Newcastle. It was not the most
powerful earthquake recorded in Australia, but it caused the
most damage.
Earthquakes produce two different types of waves: body waves
and surface waves. Body waves are seismic waves that travel
through the Earth, and can be primary or P waves (longitudinal
waves) or secondary or S waves (transverse waves), as shown in
Figure 7.2.4. P waves travel faster than S waves and therefore are
felt and detected first.
Surface waves are seismic waves that pass along the surface
of the Earth. Surface waves are usually produced by shallow
earthquakes. They have a slower speed than body waves but a
larger amplitude. They tend to do more damage than body waves.
The speed of seismic waves depends on factors such as the
density and the elasticity of the Earth. The Earth’s crust, mantle
and core are not uniform, so the speed of seismic waves can
range from 2 to 8 km s−1 in the crust to 13 km s−1 in the core.
FIGURE 7.2.4 Seismic activity produces both primary (longitudinal)
and secondary (transverse) waves.
CHAPTER 7 | WAVE PROPERTIES
225
7.2 Review
SUMMARY
• Waves can be represented by displacement–
distance graphs and displacement–time graphs.
• The period of a wave has an inverse relationship
to the frequency:
• From a displacement–time graph, you can
determine the amplitude, frequency and period.
T = 1f
• The speed of a wave can be calculated using the
wave equation:
v = f λ = Tλ
• From a displacement–distance graph, you
can determine the amplitude, wavelength and
wavenumber.
• The wavenumber of a wave has an inverse
relationship to the wavelength:
k=
2π
λ
KEY QUESTIONS
1
Consider the displacement–distance graph below.
a Which two points on the wave are in phase?
b What is the name for the distance between these
two points?
c Which two particles are at their maximum
displacement from their rest position?
d What is the term for this maximum displacement?
B
A
4
Five wavelengths of a wave pass a point each second.
The amplitude is 0.3 m and the distance between
successive crests of the waves is 1.3 m. What is the
speed of the wave?
5
Which of the following is true and which is false? For
the false statements rewrite them to make them true.
a The frequency of a wave is inversely proportional to
its wavelength.
b The period of a wave is inversely proportional to
its wavelength.
c The amplitude of a wave is not related to its speed.
d Only the wavelength of a wave determines its
speed.
6
Consider the displacement–distance graph below.
F
C
E
G
D
Use the graph below to determine the wavelength,
wavenumber and the amplitude of the wave.
Particle
displacement
(cm)
20
10
0
−10
−20
226
0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2
0.1
0
−0.1
0.4
0.6
Time (s)
MODULE 3 | WAVES AND THERMODYNAMICS
0.8
1
2
3
4
5
6
7
8
Distance (cm)
a State the wavelength and amplitude of the wave.
b If the wave moves through one wavelength in 2 s,
what is the speed of the wave?
c If the wave is moving to the right, which of the
coloured particles is moving down?
7
0.2
0
−0.5
Distance (m)
The graph below is the displacement–time graph for a
particle P.
a What is the period of the wave?
b What is the frequency of the wave?
Displacement
of P (m)
3
0.5
Displacement
(cm)
2
Calculate the period of a wave with frequency
2 × 105 Hz.
Chapter review
KEY TERMS
amplitude
compression
crest
frequency
longitudinal
mechanical wave
medium
period
phase
pulse
rarefaction
sinusoidal
transverse
trough
wavelength
wavenumber
KEY QUESTIONS
2
A mechanical wave may be described as either
transverse or longitudinal. Describe the similarities and
differences between transverse and longitudinal waves.
3
Using ideas about the movement of particles in air,
explain how you know that sound waves carry only
energy and not matter from one place to another.
4
In a transverse wave, how does the motion of the
particles compare with the direction of energy travel of
the wave?
5
Classify each of the waves described below as either
longitudinal or transverse:
a sound waves
b a vibrating guitar string
c slinky moved with an upward pulse
d slinky pushed forwards and backwards.
6
Explain why mechanical waves generally travel faster
in solids than in gases.
7
In the picture below, a circular object is creating waves.
Describe the direction of energy transfer between the
object and point X. Justify your answer.
X
8
At the moment in time shown on the graph below, in
what directions are the particles U and V moving?
V
Particle displacement (cm)
Imagine that you watch from above as a stone is
dropped into water. Describe the movement of the
particles on the surface of the water.
U
Distance (m)
9
The displacement–time graph below shows the motion
of a particular point P on a string as a wave passes
through the string. Use the graph to find the period
and frequency of the wave.
Displacement of P (m)
1
20
40
Time (ms)
60
80
10 A bee’s wings can vibrate at up to 200 Hz. What is the
period of vibration at this frequency? Give your answer
in milliseconds.
11 The distance between five successive crests on a slinky
spring is 20 cm. What is the average wavelength of
these waves?
CHAPTER 7 | WAVE PROPERTIES
227
CHAPTER REVIEW CONTINUED
The following information relates to Questions 12 and 13.
The two graphs below depict transverse waves. The scales
on both sets of axes are identical.
y
y
Wave A
15 A wave source in a ripple tank vibrates at a frequency
of 10.0 Hz. If the wave crests formed are 30.0 mm
apart, what is the speed of the waves (in m s−1) in
the tank?
16 A wave is travelling at 1200 m s−1 with a frequency of
22 kHz. What is its wavelength?
Wave B
17 Calculate the wavelength of a 300 Hz wave travelling at
1500 m s−1.
x
x
12 Calculate the ratio of the amplitude of wave A to the
amplitude of wave B.
13 Calculate the ratio of the wavelength of wave A to the
wavelength of wave B.
14 Consider the following displacement–time graph for
a sound wave. Which of the points A, B, C and D are
in phase?
C
Displacement of P (m)
A
228
20
40
Time (ms)
B
60
80
D
MODULE 3 | WAVES AND THERMODYNAMICS
18 The wave in Question 17 moves into a different
medium where its speed is 340 m s−1. What is its
wavelength now?
19 Consider two waves travelling in a medium. Wave A has
a frequency of 250 Hz, while wave B has a frequency of
1.0 kHz. What is the ratio of the wavelength of A to the
wavelength of B?
20 If you decreased the frequency of waves produced in
water, what effect would this have on the wavelength
and the velocity of the waves?
21 After completing the activity on page 216, reflect on
the inquiry question:
What are the properties of all waves and wave motion?
In your response, discuss different types of waves
(including both mechanical and electromagnetic) and
describe quantities such as wavelength and frequency.
CHAPTER
Wave behaviour
Waves, whether they are mechanical waves such as sound, or electromagnetic waves
such as light, exhibit a number of key behaviours. As scientists came to understand
and quantify these behaviours, it allowed the development of more advanced optical
systems and the refinement of musical instruments and sound systems, and also
allowed us to understand the significance of oscillating systems and the implications
of resonance.
This chapter looks at the wave behaviours of reflection, refraction, diffraction,
superposition and resonance.
Content
INQUIRY QUESTION
How do waves behave?
By the end of this chapter you will be able to:
• explain the behaviour of waves in a variety of situations by investigating the
phenomena of:
-- reflection
-- refraction
-- diffraction
-- wave superposition (ACSPH071, ACSPH072)
• conduct an investigation to distinguish between progressive and standing waves
(ASCSPH072)
• conduct an investigation to explore resonance in mechanical systems and the
relationships between: CCT
-- driving frequency
-- natural frequency of the oscillating system
-- amplitude of motion
-- transfer/transformation of energy within the system (ACSPH073) ICT N
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
8.1 Wave interactions
FIGURE 8.1.1 The reflection of waves in a ripple
tank when meeting a solid surface, in this case
a barrier positioned below the source of the
circular waves.
Mechanical waves transfer energy through a medium. A medium is necessary, but
there are times when that medium physically ends, such as when a water wave meets
the edge of a pool or air meets a wall. A change in the physical characteristics in the
same medium, such as its density or temperature, can act like a change in medium.
When the medium ends, or changes, the wave does not just stop. Instead, the energy
that the wave is carrying undergoes three processes:
• some energy is reflected (Figure 8.1.1)
• some energy is absorbed by the new medium
• some energy is transmitted.
Rarely in the real world does one mechanical wave occur in isolation. From the
ripples that form on a pond hit by raindrops to the complex interactions of multiple
reflected sound waves, the world is full of mechanical waves. The sounds produced
by acoustic musical instruments and the human voice come from the interaction
between sound waves and their reflections. The interaction of mechanical waves
results in superposition and creates, among many other things, the characteristic
sounds of musical instruments and of the human voice.
REFLECTION
When a transverse wave pulse reaches a hard surface, such as the fixed end of a
rope, the wave is bounced back or reflected.
When the end of the rope is fixed, the reflected pulse is inverted (Figure 8.1.2).
So, for example, a wave crest would be reflected as a trough.
(a)
(b)
(c)
FIGURE 8.1.2 (a) A wave pulse moves along a string towards the right and approaches a fixed post.
(b) On reaching the end, the string exerts an upwards force on the fixed post. Due to Newton’s third
law, the fixed post exerts an equal and opposite force on the string which (c) inverts the wave pulse
and sends its reflection back towards the left on the bottom side of the string. There is a phase
reversal on reflection from a fixed end.
This inversion can also be referred to as a 180° change of phase or, expressed in
terms of the wavelength, λ, a shift in phase of λ2 .
When a wave pulse hits the end of the rope that is free to move (known as a free
boundary), the pulse returns with no change of phase (Figure 8.1.3). That is, the
reflected pulse is the same as the incident pulse. A crest is reflected as a crest and a
trough is reflected as a trough.
(a)
(b)
(c)
FIGURE 8.1.3 (a) A wave pulse moves along a string towards the right and approaches a free end
at the post. (b) On reaching the post the free end of the string is free to slide up the post.
(c) No inversion happens and the wave pulse is reflected back towards the left on the same side of
the string, i.e. there is no phase reversal on reflection from a free end.
230
MODULE 3 | WAVES AND THERMODYNAMICS
When the transverse wave pulse is reflected, the amplitude of the reflected wave
is not quite the same as the original. Part of the energy of the wave is absorbed by
the post, where some will be transformed into heat energy and some will continue to
travel through the post. You can see this more clearly by connecting a heavier rope
to a lighter rope. The change in density has the same effect as a change in medium
(Figure 8.1.4).
When a transverse wave pulse is sent down the rope from the light rope to the
heavier rope, part of the wave pulse will be reflected and part of it will be transmitted
to the heavier rope. As the second rope is heavier, a smaller proportion of the wave
is transmitted into it and a larger proportion of the wave is reflected back.
This is just the same as a wave pulse striking a wall. The more rigid and/or
dense the wall is, the more the wave energy will be reflected and the less it will be
absorbed — but some energy is always absorbed by, or transferred to, the second
medium. This explains why sound can travel through walls.
light
section
(a)
heavy
section
transmitted
pulse
(b)
reflected
(a) A wave pulse travels along
a light rope towards a heavier rope. (b) On
reaching a change in density the wave pulse is
partly reflected and partly transmitted. This is
analogous to a change in medium.
FIGURE 8.1.4
REFLECTED WAVE FRONTS
(a)
(b)
ray
fronts
ray
wave fronts
wave
Two-dimensional and three-dimensional waves, such as water waves, travel as wave
fronts. When drawing wave fronts (Figure 8.1.5), it is common to show the crests
of the waves. When close to the source, wave fronts can show considerable curvature
or may even be spherical when generated in three dimensions. Where a wave has
travelled a long distance from its source, the wave front is nearly straight and is
called a plane wave. A plane wave is shown in Figure 8.1.5b. Plane waves can also
be generated by a long, flat source such as those often used in a ripple tank.
The direction of motion of any wave front can be represented by a line
drawn perpendicular to the wave front and in the direction the wave is moving
(Figure 8.1.5a). This is called a ray. Rays can be used to study or illustrate the
properties of two- and three-dimensional waves without the need to draw individual
wave fronts.
By using rays to illustrate the path of a wave front reflecting from a surface, we
can see that, for a two-dimensional or three-dimensional wave, the angle from the
normal at which the wave strikes a surface is equal to the angle from the normal to
the reflected wave. The normal is an imaginary line at 90°, i.e. perpendicular, to
the surface.
These angles of the incident and reflected waves from the normal are labelled
i and r, respectively, in Figure 8.1.6.
ray
ray
FIGURE 8.1.5 Rays can be used to illustrate the
direction of motion of a wave. They are drawn
perpendicular to the wave front of a two- or
three-dimensional wave and in the direction of
travel of the wave; (a) shows rays for circular
waves near a point source while (b) shows a ray
for plane waves.
normal
incident ray
reflected ray
i
incident
wave front
r
reflected
wave front
i
r
FIGURE 8.1.6 The law of reflection. The angle between the incident ray and the normal (i) is the same
as the angle between the normal and the reflected ray (r).
CHAPTER 8 | WAVE BEHAVIOUR
231
FIGURE 8.1.7 Reflection from an irregular
surface. Each incident ray may be reflected
in a different direction, depending upon how
rough or irregular the reflecting surface is. The
resulting wave is diffuse (spread out).
incident medium
B
v1 Δt
i
A
r
D
v2 Δt
C
transmitting medium
FIGURE 8.1.8 Wave refraction occurs because
the distance A–C travelled by the wave in the
transmitting medium is shorter than the distance
B–D that it travels at the same time in the
incident medium.
This is referred to as the law of reflection. The law of reflection states that the
angle of reflection, measured from the normal, equals the angle of incidence
measured from the normal; that is i = r.
The law of reflection is true for any surface, whether it is straight, curved or
irregular. For all surfaces, including curved or irregular surfaces, the normal
is drawn perpendicular to the surface at the point of contact of the incident ray
or rays.
When wave fronts meet an irregular, rough surface, the resulting reflection can
be spread over a broad area. This is because each point on the surface may reflect the
portion of the wave front reaching it in a different direction, as seen in Figure 8.1.7.
This is referred to as diffuse reflection.
REFRACTION
Refraction is a change in the direction of a wave caused by a change in its speed.
Changes in the speed of a wave occur when the wave passes from one medium
(substance) into another.
Consider Figure 8.1.8, where waves are moving from an incident medium where
they have high speed, v1 , into a transmitting medium where they have a lower speed,
v2 . At the same time that the wave travels a distance v1Δt (i.e. B→D) in the incident
medium, it travels a shorter distance v2Δt (A→C) in the transmitting medium. In
order to do this, the wave fronts must change direction or ‘refract’, as shown.
The direction of the refraction depends on whether the waves speed up or slow
down when they move into the new medium. In Figure 8.1.9, the water waves slow
down as they move from deeper to shallower waters, so the direction of propagation
of the wave is refracted towards the normal. The angle of incidence i, which is
defined as the angle between the direction of propagation and the normal, is greater
than the angle of refraction r.
deep
v1 > v2
λ1 > λ2
f1 = f2
i>r
original
direction
of waves
shallow
r
normal to the surface
i
λ2
λ1
wave fronts
v1
FIGURE 8.1.9
v2
Waves refract towards the normal when they slow down.
Conversely, when a wave moves from shallow waters (where it has low speed)
into deep water where it travels more quickly, it is refracted away from the normal,
as shown in Figure 8.1.10. In other words, the angle of incidence i, is less than the
angle of refraction, r.
232
MODULE 3 | WAVES AND THERMODYNAMICS
shallow
v1 < v2
deep
λ1 < λ2
direction of
refracted waves
i<r
original
direction
of waves
f1 = f2
normal to the surface
r
i
λ2
λ1
wave fronts
v1
FIGURE 8.1.10
v2
Waves refract away from the normal when they speed up.
Note that when a wave changes its speed, its wavelength also changes
correspondingly, but its frequency does not change as there is still the same number
of waves; waves cannot be gained or lost.
DIFFRACTION
When a plane (straight) wave passes through a narrow opening, it bends. Waves also
bend as they travel around obstacles (Figure 8.1.11). This ‘bending’ phenomenon
is known as diffraction.
Diffraction is significant when the size of the opening or obstacle is similar to or
smaller than the wavelength of the wave.
FIGURE 8.1.11 Water waves bend around an
obstacle. Sound waves diffract as well, allowing
you to hear around corners.
Diffraction and slit width
In the diffraction of waves, if the wavelength is much smaller than the gap or
obstacle, the degree of diffraction is less. For example, Figure 8.1.12 shows the
diffraction of water waves in a ripple tank. In Figure 8.1.12a, the gap is similar
in size to the wavelength (λ ≈ w), so there is significant diffraction and the waves
emerge as circular waves, where λ is the wavelength of the wave and w is the width
of the gap. In Figure 8.1.12b, the gap is much bigger than the wavelength (λ << w),
so diffraction only occurs at the edges.
(a)
(b)
FIGURE 8.1.12 The diffraction of water waves in a ripple tank. (a) Significant diffraction occurs
when the wavelength approximates the slit width, i.e. λ ≈ w. (b) As the gap increases, the diffraction
becomes less obvious since λ << w, but it is still present.
CHAPTER 8 | WAVE BEHAVIOUR
233
Wavelengths comparable to or larger than the diameter of the obstacle or gap
produce significant diffraction. This can be expressed as the ratio wλ ≥ 1.
WAVE SUPERPOSITION
(a)
direction
of travel
direction of travel
(b)
(c)
FIGURE 8.1.13 (a) Superposition occurs
as two wave pulses approach each other.
(b) Constructive superposition occurring.
(c) After the interaction, the pulses continue
unaltered; they do not permanently affect
each other.
Imagine two transverse mechanical waves travelling towards each other along a
string, as shown in Figure 8.1.13a. When the crest of one wave coincides with the
crest of the other, the resulting displacement of the string is the vector sum of the two
individual displacements (Figure 8.1.13b). The amplitude at this point is increased
and the shape of the string resembles a combination of the two pulses. After they
interact, the two pulses continue unaltered (Figure 8.1.13c). The resulting pattern
is a consequence of the principle of superposition. Superposition occurs when
multiple waves interact and their amplitudes are added together. In this case, as the
two waves are added together, constructive superposition occurs.
When a pulse with a positive displacement meets one with a negative
displacement as shown in Figure 8.1.14, the resulting wave has a smaller amplitude
(Figure 8.1.14b). Once again the resulting displacement of the string is the vector
sum of the two individual displacements; in this case, a negative displacement adds
to a positive displacement to produce a smaller wave. This is called destructive
superposition. Once again, the pulses emerge from the interaction unaltered
(Figure 8.1.14c).
(a)
direction of travel
direction of travel
(b)
(c)
FIGURE 8.1.14 (a) As two wave pulses approach each other, superposition occurs. (b) Superposition
of waves in a string showing destructive superposition. (c) As in constructive superposition, the waves
do not permanently affect each other.
FIGURE 8.1.15 The ripples from raindrops
striking the surface of a pond behave
independently, whether they cross each other
or not. Where the ripples meet, a complex
wave is seen as the result of the superposition
of the component waves. After interacting, the
component waves continue unaltered.
234
When two waves meet and combine, there are places where constructive
superposition occurs and places where destructive superposition occurs. Although
the wave pulses interact when they meet, passing through each other does not
permanently alter the shape, amplitude or speed of either pulse. Just like transverse
waves, longitudinal waves are also superimposed as they interact.
The effects of superposition can be seen in many everyday examples. The ripples
in the pond in Figure 8.1.15 were caused by raindrops hitting the pond. Where two
ripples meet, a complex wave results from the superposition of the two waves, after
which the ripples continue unaltered.
MODULE 3 | WAVES AND THERMODYNAMICS
Imagine two waves, one of which (Figure 8.1.16b) is twice the frequency of the
other (Figure 8.1.16a). The two individual waves are added together to give a more
complicated resultant wave, as shown in Figure 8.1.16c.
Worked example 8.1.1
(a)
frequency = f
amplitude = a
WAVE SUPERPOSITION
(b)
Two wave pulses are travelling towards each other at 1 m s−1, as shown in the
figure below.
y
frequency = 2f
amplitude = 12 a
(c)
resultant sound
wave when
superimposed
frequency = f
x
What is the amplitude of the combined pulse when they pass each other in
3 seconds?
Thinking
Working
Draw a diagram
of the two
pulses after
3 seconds.
y
FIGURE 8.1.16 Two waves, one twice the
frequency of the other, produce a complex
wave of varying amplitude when they are
superimposed.
x
Draw a new
diagram with
the waves
superimposed.
y
x
Calculate the
height of the
resultant wave.
The amplitude of the resultant wave is 1 unit.
CHAPTER 8 | WAVE BEHAVIOUR
235
Worked example: Try yourself 8.1.1
WAVE SUPERPOSITION
Two wave pulses are travelling towards each other at 1 m s–1, as shown in the
figure below.
y
x
What is the amplitude of the combined pulse when they pass each other in
2 seconds?
PHYSICSFILE ICT
Noise-cancelling headphones
Sound behaves as a mechanical wave, which you will learn about in Chapter 9. Because
of this, the principle of superposition can be applied to sound waves.
Noise-cancelling headphones utilise superposition to help reduce unwanted ambient
noise. The circuitry in the headphones analyses the ambient noise, and produces a new
sound (wave form) to counteract the unwanted sound. The newly produced sounds are
inversions (out of phase by 180°) of the unwanted sound, so that when the two sounds
are added together (superimposed) they cancel each other out.
(a)
(b)
y
y
x
x
(c)
(d)
y
x
y
x
FIGURE 8.1.17 (a) The ambient sound waves detected by the microphone in the headphone.
(b) The new sound wave that is produced. This is an inverted version of the ambient sound.
(c) The sum of these two waves (d) cancels out the unwanted noise.
236
MODULE 3 | WAVES AND THERMODYNAMICS
STANDING WAVES
Earlier in this chapter it was shown that when a wave pulse reaches a fixed end, it is
reflected back 180° out of phase. That is, crests are reflected as troughs and troughs
are reflected as crests.
Imagine creating a series of waves in a rope by shaking it vigorously. As the rope
shakes, waves travel in both directions along it. The new waves travelling down the
rope interfere with those being reflected back. This kind of motion usually creates
quite a random pattern, with the waves quickly dying away. Shaking the rope at
just the right frequency, however, will create a new wave that interferes with the
reflection in such a way that the two superimposed waves create a single, largeramplitude wave. This type of wave is known as a standing wave.
It is called a standing wave because it does not appear to be travelling along the
rope. The rope seems to simply oscillate up and down with a fixed pattern. That is,
it seems to be just ‘flipping’ up and down in the same place. This situation contrasts
with a standard transverse wave where every point on the rope has a maximum
displacement at some time as the wave travels along the rope.
In Figure 8.1.18a to d, two waves (drawn in blue) are shown travelling in
opposite directions towards each other along a rope. One of the waves is a string of
pulses (shown as a solid line) and the other is its reflection (shown as a dotted line).
The two waves superimpose when they meet. Since the amplitude and frequency of
each is the same, the end result, shown in Figure 8.1.18e, is a standing wave. At the
points where destructive interference occurs, the two waves totally cancel each other
out and the rope remains still.
λ
resultant
direction of wave
(a)
direction of wave
resultant
(b)
(c)
resultant
resultant
(d)
(e)
λ
2
λ
2
FIGURE 8.1.18 (a)–(d) Two
waves (one a solid blue line
and the other a dotted blue
line) travelling in opposite
directions, each with
the same amplitude and
frequency, interact by the
principles of superposition
creating the resultant wave
(the dotted red line). (e) A
standing wave is created.
CHAPTER 8 | WAVE BEHAVIOUR
237
8.1 Review
SUMMARY
• A wave reaching the boundary between two
materials in which it can travel will always be
partly reflected, partly transmitted and partly
absorbed.
• Refraction is the change in direction of
propagation of a wave that occurs when there is a
change in speed of the waves.
• A wave has been reflected if it bounces back after
reaching a boundary or surface.
• When a plane (straight) wave passes through
a narrow opening or meets a sharp object, it
experiences diffraction.
• Waves reflect with a 180° phase change from fixed
boundaries. That is, crests reflect as troughs and
troughs reflect as crests.
• Significant diffraction occurs when the wavelength
of the wave is similar to, or larger than, the size of
the diffracting object.
• Waves reflect with no phase change from free
boundaries. That is, crests reflect as crests and
troughs as troughs.
• The principle of superposition states that
when two or more waves interact, the resultant
displacement or pressure at each point along the
wave is the vector sum of the displacements or
pressures of the component waves.
• When a wave is reflected from a surface, the angle
of reflection is equal to the angle of incidence.
KEY QUESTIONS
1
A wave travels along a rope and reaches a fixed end.
What occurs next?
2
Which of the following properties of a wave can change
when the wave is reflected: frequency, amplitude,
wavelength or speed?
3
Two triangular wave pulses head towards each other at
1 m s−1. Each pulse is 2 m wide.
4
A ray strikes a flat surface at an angle of 38° measured
from the surface. What is the angle of reflection of
the ray?
5
The following diagram shows a wave before and after
being reflected from an object.
incident
wavefronts
reflected
wavefronts
What will the superposition of these two pulses look
like in 3 s?
A
B
C
D
238
MODULE 3 | WAVES AND THERMODYNAMICS
What is the shape of the object?
A flat
B concave
C convex
D parabolic
6
A wave passes from medium A to medium B, as
shown in the diagram below. The wave is refracted
towards the normal.
normal
medium A
medium B
a How does the speed of the wave in medium A
compare to the speed in medium B?
b How does the frequency of the wave in medium A
compare to the frequency in medium B?
7
Consider a wave of wavelength 55 cm passing through
a gap of 50 cm as shown in the diagram below. If the
width of the opening remains unchanged but the
frequency of the wave is increased, what would be
observed? Initially, diffraction of the wave through the
gap is observed.
wavefronts
CHAPTER 8 | WAVE BEHAVIOUR
239
PHYSICS INQUIRY
CCT N
Resonant
frequency
How do waves behave?
COLLECT THIS…
•
•
•
•
steel tin
drill
string
magnet
8.2 Resonance
You may have heard about singers who supposedly can break glass by singing
particularly high notes. Figure 8.2.1 shows a glass being broken in much the same
way. All objects that can vibrate tend to do so at a specific frequency known as
their natural frequency or resonant frequency. Resonance is when an object
is exposed to vibrations at a frequency known as the driving frequency equal to
their resonant frequency. Resonance occurs when a weak vibration from one object
causes a strong vibration in another. If the amplitude of motion of the vibrations
becomes too great, the object can be destroyed.
DO THIS…
1 Drill a hole through the top
of the tin on two sides, and
thread the string through.
2 Tie the tin to a clothesline or
another high place, creating
a pendulum. Measure the
length of the pendulum.
3 Tie some string to the
magnet.
4 Stick the magnet to the
side of the tin. Pull the
string softly to start the tin
swinging.
5 Try pulling the string at
different points in the swing
to increase the amplitude.
Measure the frequency of
the pulls that create the
biggest swings.
6 Change the length of the
pendulum and repeat
steps 4 and 5.
RECORD THIS…
Describe the motion of the
pendulum as the driving frequency
is changed.
Present your results, graphing the
length vs. the natural frequency.
REFLECT ON THIS…
How do waves behave?
Where else have you seen
resonance occur?
240
FIGURE 8.2.1 A glass can be destroyed by the vibrations caused by a singer emitting a sound of the
same frequency as the resonant frequency of the glass.
A swing pushed once and left to swing or oscillate freely is an example of an
object vibrating at its natural frequency. The frequency at which it moves backwards
and forwards depends entirely on the design of the swing, mostly on the length of its
supporting ropes. In time, the oscillations will fade away as the energy is transferred
to the supporting frame and the air.
If you watch a swing in motion, it is possible to determine its natural oscillating
frequency. It is then possible to push the swing at exactly the right time so that
you match its natural oscillation. The energy you add by pushing will increase the
amplitude of the swing rather than work against it. Over time, the amplitude will
increase and the swing will go higher and higher; this is resonance. The swing can
only be pushed at one particular rate to get this increase in amplitude (i.e. to get the
swing to resonate). If the rate is faster or slower, the forcing frequency that you are
providing will not match the natural frequency of the swing and you will be fighting
against the swing rather than assisting it.
Other examples of resonant frequency that you may have encountered are
blowing air across the mouthpiece of a flute or drawing a bow across a string of a
violin in just the right place (Figure 8.2.2). In each case, a clearly amplified sound
is heard when the frequency of the forcing vibration matches a natural resonant
frequency of the instrument.
Two very significant effects occur when the natural resonant frequency of an
object is matched by the driving frequency.
• The amplitude of the oscillations within the resonating object increases
dramatically.
• The maximum possible energy from the source creating the forced vibration is
transferred to the resonating object.
MODULE 3 | WAVES AND THERMODYNAMICS
In musical instruments and loudspeakers, resonance is a desired effect. The
sounding boards of pianos and the enclosures of loudspeakers are designed to
enhance and amplify particular frequencies. In other systems, such as car exhaust
systems and suspension bridges, resonance is not always desirable, and care is taken
to design a system that prevents resonance.
PHYSICSFILE ICT CCT
Tacoma Narrows Gorge suspension bridge
One of the most recognisable cases of mechanical resonance for many years has been
the destruction of the Tacoma Narrows Gorge Bridge in the US State of Washington in
1940. It was originally
thought, from studying
video footage of the
bridge’s collapse that the
wind acted as a driving
frequency, causing the
bridge to oscillate with
ever-increasing amplitude
until the whole bridge
shook itself apart. More
recent research seems to
suggest that instead the
wind supplied a twisting
motion causing the bridge FIGURE 8.2.3 The Tacoma Narrows Gorge Bridge, Washington,
U.S.A. bends and buckles due to the force of wind.
to tear itself apart.
PHYSICS IN ACTION
WE
FIGURE 8.2.2 The sound box of a stringed
instrument is tuned to resonate for the range of
frequencies of the vibrations being produced by
the strings. When a string is plucked or bowed,
the airspace inside the box vibrates in resonance
with the natural frequency and the sound is
amplified.
ICT
Resonance in aircraft wings
Have you ever looked out of an aeroplane window and
noticed that the wings are vibrating up and down? This
effect, sometimes known as flutter, is due to the vibrational
energy both of the engines on the aeroplane and the
air flow across the wings. While small vibrations in the
wings are normal, resonance in the wings is not a desired
effect. Every effort is made to make sure that the driving
frequency of the engines and the driving frequency of
the air flow do not match the natural resonant frequency
of the wings. Aeronautical engineers do not want the
energy from the driving vibrations to be transferred to the
aeroplane wings. Engineers and pilots test aeroplanes by
flying them at great speeds, often close to the speed of
sound, to see how they manage the vibrations that such air
flow causes.
FIGURE 8.2.4 Small vibrations in aeroplane wings are expected.
Resonance in aeroplanes is not a desired effect.
CHAPTER 8 | WAVE BEHAVIOUR
241
8.2 Review
SUMMARY
• Resonance occurs when the driving frequency (a
forcing vibration) equals the natural frequency of
an object.
• Two special effects occur with resonance:
-- the amplitude of vibration increases
-- the maximum possible energy from the source
is transferred to the resonating object.
KEY QUESTIONS
Explain why resonance can result in damage to
man-made structures.
2
Resonance occurs when the driving frequency of
a vibration exactly equals the natural frequency of
vibration of an object. Two special effects occur. Which
one of the following responses relating to the effects of
resonance is true? Explain your answer.
A The amplitude of vibration will decrease.
B The amplitude of vibration will increase.
C The frequency of vibration will increase.
D The frequency of vibration will decrease.
3
The frequency of a pendulum depends on its length.
Consider a set of pendulums, all attached to a
horizontal bar as shown in the diagram below. If
the pendulum on the left is made to oscillate, which
pendulum, A, B, C or D, would oscillate with the largest
amplitude? Explain your answer.
A
C
D
B
4
242
While sitting in a stationary truck with the engine idling
at 100 Hz, the driver notices that the side mirror on the
outside of the truck is vibrating significantly. When the
truck is being driven along the road with the engine
revolutions greater than 100 Hz, they notice that the
vibrations of the mirror are no longer as significant.
Explain why this would be the case.
MODULE 3 | WAVES AND THERMODYNAMICS
5
A footbridge over a river has a natural frequency of
oscillation from side to side of approximately 1 Hz.
When pedestrians walk at a pace that will produce an
oscillation in the bridge close to its natural frequency,
resonance occurs. The graph below shows data
about pedestrians walking or running. A pedestrian
completes one cycle of their motion every two steps.
Which activity of the pedestrians is most likely to
cause damage to the footbridge over time? Explain
your answer.
Steps per second vs speed
3.5
fast running
3
Steps per second
1
2.5
2
normal
walk
slow running
fast walk
slow walk
1.5
1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5
Speed (m s−1)
Chapter review
KEY TERMS
absorb
angle of incidence
angle of reflection
diffraction
diffuse
driving frequency
natural frequency
normal
plane wave
ray
reflect
refraction
resonance
standing wave
superposition
transmit
wave front
REVIEW QUESTIONS
1
A pulse is sent along a rope that is fixed at one end.
What property of the pulse changes when it is reflected
at the fixed end of the rope?
2
When sunlight shines through a window in a house,
its energy can be transmitted, reflected or absorbed.
Which of these processes is responsible for the
fact that:
a the interior of the house is illuminated?
b when light falls on a window, you can see some of it
from outside the house?
c the glass gets warm?
3
4
Which of the following about wave pulses are true and
which are false? For the false statements, rewrite them
so they are true.
a The displacement of the resultant pulse is
equal to the sum of the displacements of the
individual pulses.
b As the pulses pass through each other, the
interaction permanently alters the characteristics of
each pulse.
c After the pulses have passed through each other,
they have the same characteristics as before
the interaction.
What phenomenon does the diagram below
demonstrate?
A diffraction
B interference
C reflection
D refraction
5
The figure represents a situation involving the
refraction of light. Identify the correct label for each
of the lines from the choices provided: boundary
between media, reflected ray, incident ray, normal,
refracted ray.
A
B
C
medium 1
medium 2
D
E
6
When arriving at a concert or a party, explain why the
bass sounds are more easily heard from a distance
than the higher-pitch sounds.
The following information relates to Questions 7 and 8:
When viewed from above, a fish in water appears to be
closer to the surface of the water than it actually is. This
is shown in the diagram: the fish is actually located at (A),
but seems to be located at (B).
eye
normal
air
water
(A)
7
(B)
Explain why the light refracts away from the normal as
it passes from the water into the air.
CHAPTER 8 | WAVE BEHAVIOUR
243
CHAPTER REVIEW CONTINUED
8
If the fish were viewed from directly above (viewing
along the normal), how would its apparent
position change?
9
When waves pass through narrow openings, diffraction
can be observed. Explain why we do not observe
diffraction of light in our everyday lives.
The following information relates to Questions 10, 11, 12
and 13.
The diagram below shows the path of sound from a rock
concert, across a lake with warm air over the lake and
hotter air above.
rock concert
hotter air
building
warm air
lake
10 What two wave behaviours are demonstrated by
this situation?
11 What does the path of the sound wave from the warm
air to the hotter air tell us about the speed of sound in
warm air compared to the speed of sound in hotter air?
12 Which of the following statements is correct regarding
the sound of the rock concert?
A The sound seems higher-pitched to the people
on top of the building than to those on the
ground floor.
B The sound seems lower-pitched to the people
on top of the building than to those on the
ground floor.
C The sound seems to be at the same pitch to the
people on top of the building and to those on the
ground floor.
13 If the hotter body of air cools until it reaches the same
temperature as the lower body of air, what implication
will that have for the people in the building?
14 The following graph shows three wave forms. Two
of the wave forms superimpose to create the third
wave form.
Which wave is the result of the superposition of the
other two?
2
1
–1
1
2
3
4
5
–2
244
MODULE 3 | WAVES AND THERMODYNAMICS
6
15 Two wave pulses are travelling along a string towards
each other in opposite directions. When they meet,
the waves superimpose and cancel each other out
completely. What characteristics must the two waves
have to make this happen?
16 Describe the concept of resonance using the terms
natural and driving frequency. Give an example of
where you may find resonance.
17 In order to produce significant diffraction of red light
(wavelength approximately 700 nm), a diffraction
experiment would need to use an opening with a width
of approximately:
A 1 mm
B 0.1 mm
C 0.01 mm
D 0.001 mm
18 A transverse standing wave is produced using a rope.
Is the standing wave actually standing still? Explain
your answer.
19 There are certain sounds (frequencies) that appear
louder to humans than other sounds. These sounds
are typically in the region of 2500 Hz. Explain why a
2500 Hz sound from a signal generator sounds louder
or ‘resonates’ well in the ear canal of humans.
20 Reflection is possible from which of the following
surface shapes if they are all reflective? (More than one
answer may be correct.)
A flat surface
B concave (curved inwards) surface
C uneven surface
D convex (curved outwards) surface
21 After completing the activity on page 240, reflect on
the inquiry question: How do waves behave? In your
response, discuss the concept of resonance.
CHAPTER
Sound waves
Sound is important to many forms of communication for humans and other
animals. For many, it is the major form of receiving information about the world
around them. As music, it is a form of entertainment, lifting the spirit and allowing a
depth of expression rivalled in few other fields.
This chapter introduces some acoustic concepts, including sound intensity and the
production of standing waves in strings and pipes.
Content
INQUIRY QUESTION
What evidence suggests that sound is a mechanical wave?
By the end of this chapter you will be able to:
• conduct a practical investigation to relate the pitch and loudness of a sound to its
wave characteristics
• model the behaviour of sound in air as a longitudinal wave
• relate the displacement of air molecules to variations in pressure (ACSPH070)
• investigate quantitatively the relationship between distance and intensity of sound
• conduct investigations to analyse the reflection, diffraction, resonance and
superposition of sound waves (ACSPH071)
• investigate and model the behaviour of standing waves on strings and/or in pipes
to relate quantitatively the fundamental and harmonic frequencies of the waves
that are produced to the physical characteristics (e.g. length, mass, tension, wave
velocity) of the medium (ACSPH072) ICT N
• analyse qualitatively and quantitatively the relationships of the wave nature of
sound to explain: CCT
-- beats fbeat =f2 − f1
v
+v
-- the Doppler effect f ′ = f ( vwave − vobserver ) .
(
wave
source
)
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
9.1 Sound as a wave
PHYSICS INQUIRY CCT N
Sound interference
What evidence suggests that sound is a mechanical wave?
COLLECT THIS…
•
two speakers, with connector
•
sound source such as frequency generator, computer, phone
•
16 cones
•
measuring tape
DO THIS…
1 Set up the speakers and sound source outside in an area that does not have many hard surfaces nearby.
2 Set the frequency generator to 340 Hz. This will produce sound waves with a wavelength of approximately 1 m.
3 Starting 2 m from the speakers, walk across in front of the speakers, placing a cone when the sound is quiet. Place
four cones, two on each side of the centre line, creating a row of cones.
4 Repeat at 4 m, 6 m and 8 m from the speakers.
5 Choose a line (or column) of cones and measure the distance from each speaker to each cone.
6 Record the path difference along with the cone number from the edge. Write the path difference as a fraction of the
wavelength.
RECORD THIS…
Describe what you heard as you walked in a straight line in front of the speakers.
Present your results, drawing a picture of the pattern produced by the cones, and any relationship found in the path
difference in terms of wavelength.
REFLECT ON THIS…
What evidence suggests that sound
is a mechanical wave?
Why were there some areas of
quiet sound but no areas where
there was no sound?
centre line
speakers
2m
FIGURE 9.1.1 Compressions and rarefactions of
a pressure wave created by the vibrations of a
tuning fork.
246
2m
2m
2m
Vibrations are responsible for all sounds that we hear. Our vocal cords vibrate in our
larynx producing our voice sounds. Drum skins vibrate when struck with sticks to
make the sounds we hear. Speakers vibrate and create the sounds we hear from our
home theatre systems or our headphones.
Tuning forks are used to tune musical instruments. If you observe the prongs
on the tuning fork closely, you will see them vibrating back and forth very rapidly.
A tuning fork produces a single note that depends on the of vibration frequency.
Figure 9.1.1 illustrates a vibrating tuning fork and the effect it has on the air
surrounding it.
MODULE 3 | WAVES AND THERMODYNAMICS
LONGITUDINAL WAVES
In Chapter 7 we saw that sound is a longitudinal wave. That is, the vibrations of
the molecules within the medium are in the same direction or parallel to the line
of travel of the wave energy (Figure 9.1.2). A sound wave consists of a series of
compressions and rarefactions.These compressions and rarefactions can be seen
in Figure 9.1.3. Compressions are regions where the particles within the medium
that the sound is travelling in are forced closer together. When sound is travelling
through air, these compressions are regions of increased air pressure. Rarefactions
are regions where the particles within the medium are pulled apart. When sound is
travelling through air, these rarefactions are regions of decreased air pressure. The
periodic variation in pressure forms what is known as a sound wave.
Sound cannot travel through a vacuum (including outer space) because a vacuum
has no molecules to transfer the vibrations from a vibrating source (speaker, tuning
fork, voice).
longitudinal wave
vibration
direction of travel
or propagation
FIGURE 9.1.2 In a longitudinal wave, the
vibration of the particles within the medium
are in the same direction as, i.e. parallel to, the
direction of energy flow of the wave.
Sound waves transfer energy through the vibration of molecules. Therefore, for
sound to travel, there must be a medium.
rest position of source
vibrating
source
compression
(high pressure region)
rarefaction
(low pressure region)
FIGURE 9.1.3 Before the source begins to vibrate, the particles in the medium are evenly spaced.
Once the source of the sound begins to move (vibrate) it causes the particles to clump together into
areas of higher density (compressions) and spread out in areas of lower density (rarefactions). This is
how sound is propagated.
CHAPTER 9 | SOUND WAVES
247
VISUALISING SOUND AS A WAVE
GO TO ➤ SkillBuilder, page 221
Drawing a diagram of sound waves as longitudinal waves is problematic, as a
longitudinal wave diagram such as in Figures 9.1.1 or 9.1.2 does not show the
amplitude of the wave. It is also difficult to accurately measure the wavelength
from such a diagram. Discerning exactly where the centre of compression is may
be challenging.
Because of this, physicists have adopted a series of conventions that allow sound
waves (longitudinal waves) to be represented as transverse waves.
Figure 9.1.4 is a simple representation of a transverse wave. In Chapter 7,
the characteristics of waves (amplitude, wavelength, period and frequency)
were explained.
λ
crest or peak
amplitude
D
C
A
B
amplitude
particle
motion
direction of motion
of wave
trough
λ
FIGURE 9.1.4 It can be helpful to view sound waves as transverse waves. Both A-B and C-D represent
one wavelength.
These characteristics apply to all waves, both longitudinal and transverse. If we
consider the longitudinal sound wave as shown in Figure 9.1.5a, we see the repeating
pattern of compression and rarefaction in the air in front of the vibrating speaker.
Points W and X are located at adjacent compressions. These two points are in
phase and this means that they are one wavelength apart. Conversely, points Y and Z
are one wavelength apart as they are located at adjacent rarefactions.This information
can be depicted as a wave similar to the transverse wave in Figure 9.1.4. A graph of
pressure variation vs the distance from the source at a given time can be drawn. The
compressions (increased air pressure) are depicted as peaks and rarefactions (decreased
air pressure) are shown as troughs. This type of graph is shown in Figure 9.1.5b.
(a) loudspeaker diaphragm
vibrations
C = compression
C
loudspeaker
(b)
pressure variation
W
W
R = rarefaction
R
C
R
Y
X
Z
direction of travel of sound waves
λ
C
R
longitudinal displacement
of air molecules
maximum pressure
X
a
undisturbed air pressure
distance
Y
C
FIGURE 9.1.5
248
R
Z
C
R
minimum pressure
The compressions of a longitudinal wave coincide with the peaks of the transverse wave representation.
MODULE 3 | WAVES AND THERMODYNAMICS
The wavelength of the longitudinal wave is easily measured. The frequency, f,
describes the number of wavelengths to pass a point each second. A louder sound is
produced by a vibrating source with more energy causing a larger pressure variation.
Therefore it is represented by a wave with a greater amplitude.
SOUND WAVE BEHAVIOUR
As sound is a wave, it behaves as other waves do. In Chapter 8, wave behaviour such
as reflection, diffraction, resonance and superposition were explained. Some further
examples of sound wave behaviour are given below.
GO TO ➤ Chapter 8, page 230
Reflection
The reflection of sound waves is a commonly observed phenomenon and is known
as an echo. Sound waves follow the law of reflection, in which the angle of incidence
is equal to the angle of reflection. Sound reflection is used by some animals, for
instance bats and dolphins, for echo location.
Ultrasound images of babies in the womb are generated by the reflection of highfrequency sound waves. A typical ultrasound image is shown in Figure 9.1.6. These
waves are reflected back to the receiver when they strike an object, and a computer
connected to the receiver creates an image from the signals. The longer it takes for a
signal to return to the receiver, the further away the object is located.
WS
3.4
FIGURE 9.1.6
An ultrasound image of a baby in the womb.
Diffraction
Diffraction, discussed in Chapter 8, is the spreading-out of a wave as it passes the
edge of an object, or when it has passed through a gap. Significant diffraction occurs
when the wavelength is approximately equal to or larger than the object or the gap
width. The diffraction of sound waves is relatively easy to observe. High-pitched
sound waves have short wavelengths, and therefore do not diffract as much as
low-pitched sounds which have larger wavelengths.
GO TO ➤ Section 8.1, page 233
CHAPTER 9 | SOUND WAVES
249
Generally speaking, higher-pitched (shorter-wavelength) sounds tend to be more
directional (are diffracted less) than lower-pitched, larger-wavelength sounds that
are diffracted more readily around objects and through apertures (Figure 9.1.7).
(a)
(b)
diffracted waves
diffracted waves
undiffracted
waves
source of sound
source of sound
barrier
barrier
The amount a wave will be diffracted when passing through a gap depends on both the wavelength and the size of the aperture. The
amount of diffraction (how much the wave bends) is (a) small if the size of the gap is much larger than the wavelength, compared to (b) where the
gap is smaller than the wavelength.
FIGURE 9.1.7
9.1 Review
SUMMARY
• Sound waves transfer energy by the vibration of
molecules. For sound to travel there must be a
medium.
• The changes in pressure within the medium
create a pattern of compressions (high pressure)
and rarefactions (low pressure). Sound waves can
be modelled using a pressure vs distance graph.
• Sound consists of longitudinal waves—that is, the
oscillations are parallel to the direction the wave
energy is travelling.
KEY QUESTIONS
1
250
The vibrating strings in a guitar produce sound. Which
of the following statements is correct?
A The motion of the strings creates kinetic energy. This
energy is transmitted instantaneously to your ears.
B The motion of the strings causes nearby air molecules
to vibrate, transferring energy through the medium by
a chain reaction before reaching your ears.
C The motion of the strings causes nearby air
molecules to be pushed outwards, eventually hitting
your ears.
D The motion of the strings causes a transverse wave
to propagate through the medium.
MODULE 3 | WAVES AND THERMODYNAMICS
2
Which one or more of the following statements is true?
A Sound is not able to travel through a vacuum.
B Once a sound wave has passed through a medium
the molecules do not return to their original
positions.
C Sound is a transverse wave.
D Sound waves transfer energy.
3 The figure below shows a tuning
fork creating compressions and
rarefactions in air.
P
Q
R
S
T
0
0.20
0.40
0.60
0.80
X (m)
a In what direction are the air
molecules vibrating?
A left
B right
C up
D down
E left and right
F up and down
b Calculate the wavelength.
c Sketch a pressure variation vs
distance graph of the sound
wave. (No scale is required on
the pressure variation axis.)
A detector is set up some
distance away from a sound
source which produces
the following pressure vs
time graph.
Pressure variation
(arbitrary units)
4
2.0
1.0
0
1.0
2.0
3.0
4.0
5.0
6.0
–1.0
–2.0
t (ms)
a When does a compression reach the detector?
b What is the frequency of the wave?
CHAPTER 9 | SOUND WAVES
251
9.2 Sound behaviour
We have seen how sounds are produced by vibrating objects. Musical instruments
produce sounds by vibrations. Vibration of our vocal cords allows us to speak.
Vibrations vary in frequency and amplitude. An increase in the amplitude of
vibration is produced by increasing the energy supplied. We perceive these changes
in the vibrations as changes in volume and pitch of the sound.
It is difficult to ‘see’ a sound wave, so physicists digitise sounds using a
microphone connected to a cathode ray oscilloscope (CRO) or to a computer with
the appropriate software. Changes in air pressure change the induced voltage in the
microphone, which in turn are displayed on a screen, allowing us to visualise sound
wave properties such as amplitude and frequency.
The centre image in Figure 9.2.1 shows the pattern produced when a
loudspeaker connected to a signal generator is placed near a microphone connected
to a CRO or computer. The single tone of constant volume and constant pitch
produces a uniform function (trace) that looks like a transverse wave, but we know
that sound consists of longitudinal waves. The crests and troughs in the CRO or
screen image correspond to the compressions and rarefactions caused by the sound
wave. The trace is a graph of pressure variation vs time at the microphone position.
increased
amplitude
0.0
0.005
Time (s)
0.010
0.0
decreased
amplitude
0.0
0.005
Time (s)
increased
frequency
0.010
0.0
0.005
Time (s)
original trace
0.010
0.005
0.010
0.005
0.010
Time (s)
decreased
frequency
0.0
Time (s)
FIGURE 9.2.1 Increased loudness means increased wave energy, which results in increased amplitude. Higher frequency means that
more wavelengths pass a point every second, and is heard as a higher pitch.
A single tone is shown in the centre diagram in Figure 9.2.1. This note is of
a fixed pitch (frequency) and a fixed volume or loudness (amplitude). The other
diagrams in Figure 9.2.1 illustrate how changes in volume and pitch affect the trace.
More energy in the sound wave creates a louder sound (i.e. increased amplitude).
Frequency is the number of wavelengths that pass a point in one second, so a higher
frequency of sound has a shorter wavelength and is heard as a higher pitch.
• The pitch of a sound depends on its frequency. An increase in frequency (or
a decrease in wavelength, as they are inversely related) is perceived as an
increase in pitch.
• The volume of a sound depends on its amplitude. An increase in amplitude is
perceived as an increase in volume.
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MODULE 3 | WAVES AND THERMODYNAMICS
INTENSITY
Intensity is a measure of the rate at which energy passes through an area of space
each second. The intensity of a sound wave is roughly equivalent to the volume of
the sound. Intensity is measured in watts per square metre (W m−2).
TABLE 9.2.1
1 W m−2 is equivalent to 1 J s−1 m−2.
Sound intensity levels from different sources.
PHYSICSFILE N
Intensity (W m−2)
Source
Hearing range
10−12
Lowest level of sound heard by humans
The average young person can hear
frequencies in the range of 20 Hz to
approximately 20 kHz. Sounds with
a frequency greater than 20 kHz are
known as ultrasonic. Sounds that have
a frequency less than 20 Hz are known
as infrasonic. The human voice has
frequencies in the range of 85 Hz to
1100 Hz. The range of sounds produced
and detected may be vastly different
in other animals. Bats, for instance,
navigate using high-frequency ultrasonic
sounds well outside our hearing range.
−5
Traffic
10
−2
Fire alarm
10
Pain threshold
1
Intensity (I) describes the rate of energy passing through an area each second.
The definition of intensity and its units (W m−2) lead us to express intensity as
.
I = power
area
Wave energy such as light and sound spreads out in three dimensions. Therefore,
the area that the wave energy passes through at a distance r from the source is given
as the surface area of a sphere 4πr2. The intensity is then defined as:
I=
Sounds
P
4 π r2
where: I is the intensity (in W m−2)
P is the power (in W)
r is distance from the source (in m).
human
produced
85–1100
sound
source
FIGURE 9.2.3
r
7000–120000
150–150000
452–1080
15–50000
dog
cat
760–1520
60–65000
10000–120000
1000–120000
bat
3r
2r
20–20000
dolphin
If we consider a sound source such as a speaker, and simplify this scenario
by considering the speaker to be a point source of sound, then the sound will
spread out radially, both vertically and horizontally (Figure 9.2.3). If you double
the distance from the source (from r to 2r), the sound energy is now spread
over twice the distance in two dimensions, or it has spread out over an area four
times the size at distance r. Spreading the energy over a larger area decreases the
intensity proportionally—that is, the intensity decreases by a factor of four.
heard
0
100
1000
10000
Frequency (Hz)
100000
FIGURE 9.2.2 Different frequencies of
sound that humans and some animals can
produce and detect.
The intensity (energy) of sound follows an inverse square law.
This pattern between intensity and distance is referred to as the inverse square
law. Sound intensity, I, decreases with the square of the distance, r, from a
point source.
I ∝ r12
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253
The inverse square law can only be applied if we assume that no other energy is
lost for some reason. For example, wind and temperature differences may affect the
way energy is transmitted through the medium.
Worked example 9.2.1
THE INVERSE SQUARE LAW AND INTENSITY
SKILLBUILDER
N
Understanding
inverse and
inverse square
relationships
Some relationships in science
involve one quantity in a
relationship increasing and
the other quantity decreasing
proportionally. This is an inverse
relationship.
For example, rearranging the
wave equation v = f × λ to f = λv ,
or f ∝ 1λ . In other words, as the
wavelength of a wave travelling
at constant speed doubles, the
frequency will halve.
An inverse square relationship is
similar, but one of the quantities
increases as the square of the
other quantity decreases. The
intensity of a sound wave is
directly related to the inverse
square of the distance from the
source. As the distance from the
source doubles, the intensity
decreases by a factor of 22. As the
distance from the source triples,
the intensity decreases by a
factor of 32. This is written as the
relationship I ∝ 12 .
r
254
At a concert, speakers at the front of the stage 1 m from where you are standing
produce sound with an intensity of 2.0 × 10−2 W m−2. You move further from the
stage until the intensity falls to 3.2 × 10−5 W m−2.
How far from the stage do you end up?
Thinking
Working
Write out the inverse square law.
I∝
1
r2
Create a relationship which compares
the two intensities (I1 and I2) and the
two distances (r1 and r2)
I1 ∝
1
r12
I2 ∝
1
r22
I
r22
2
r12
∴ I1 ∝
Substitute the known values and solve
for r2.
I1
I2
.
r22
∝
r12
2.0 × 10−2
3.2 × 10
−5
2
2
=
r =
r22
1.02
2 × 10−2
3.2 × 10−5
× 1.0
= 625
r2 = 25 m
Worked example: Try yourself 9.2.1
THE INVERSE SQUARE LAW AND INTENSITY
The sound from a drill is 10 mW m−2 at a distance of 1.0 m. Assume that the
sound spreads out equally in all directions.
Calculate the intensity at a distance of 10 m.
THE DOPPLER EFFECT
The Doppler effect is a phenomenon of waves that is observed whenever there is
relative movement between the source of the waves and an observer. It causes an
apparent increase in frequency when the relative movement is towards the observer
(i.e. the distance between observer and wave source is decreasing) and an apparent
decrease in frequency when the relative movement is away from the observer (i.e.
the distance between observer and wave source is increasing). It can be observed for
any type of wave, and it has been particularly useful in astronomy for understanding
the expanding universe.
The Doppler effect is named after the Austrian physicist Christian Doppler,
who proposed it in 1842 to explain the apparent change in the frequency of the
wave. The actual frequency of the wave source or signal does not change. You may
have experienced the Doppler effect, for instance, when hearing the sound of the
siren on an emergency vehicle as it approaches and passes by.
If a wave source, such as an ambulance siren, is travelling at the same speed as
you (including when both are stationary), you will receive and hear the disturbances
(rarefactions and compressions in the case of sound waves) at the same rate as the
source creates them. If the wave source travels towards you, then each consecutive
MODULE 3 | WAVES AND THERMODYNAMICS
disturbance originates from a position a little closer than the previous one, and
therefore has a slightly shorter distance to travel before reaching you than the one
immediately before it.The effect is that the frequency of arrival of the disturbances—
that is, the pitch of the sound—is higher than the originating frequency (the person
on the right in Figure 9.2.4).
stationary object
AMBULANCE
moving object
AMBULANCE
PHYSICSFILE ICT
The sound of the Doppler
effect
The Doppler effect behaviour can be
easily modelled. You should be able
to mimic the sound of a high-powered
racing car like that in Figure 9.2.5 by
making the sound ‘eee…owwwww’
with your voice. The ‘eee’ is the
sound the car makes as it approaches
you—hence the high frequency. The
‘owwwww’ is the sound it makes after
it passes you and is travelling away—
hence the low frequency.
FIGURE 9.2.4 The Doppler effect. An object emitting a sound moving towards the person on the right
emits sound waves closer together in its direction of travel and hence they hear a higher-frequency
sound. As it moves away from the person on the left, the sound waves reach the observer further
apart and are heard as a lower-frequency (deeper) sound.
Conversely, if the source is moving away from the observer, each consecutive
disturbance originates from a distance a little further away than the one immediately
before and so travels a greater distance before it reaches the observer. The
disturbances are therefore heard by the observer at a lower frequency than the
originating frequency (the person on the left in Figure 9.2.4).
The net effect is that when the wave is moving towards an observer, the frequency
of arrival of the wave is higher than the frequency of the original source. When the
wave is moving away from the observer, the frequency of arrival will be lower than
the frequency of the original source. So the Doppler effect makes the siren’s sound
appear to be at a higher frequency as the vehicle is travelling towards you, but at a
lower frequency as the vehicle moves away from you.
For a mechanical wave, the total Doppler effect may result from the motion of
the source, the motion of the observer, or the motion of the medium the wave travels
through. For waves that don’t require a medium, such as light, only the relative
difference in speed between the observer and the source contributes to the effect.
FIGURE 9.2.5 Daniel Ricciardo of Australia
racing in Spain.
Doppler calculations
As the relative motion between the observer and source is the cause of the change
in apparent frequency, then by knowing what the relative motion is, the apparent
frequency can be calculated. In classical physics (which does not take into account
relativistic effects), where the speeds of the source and the observer are less than the
speed of the waves in the medium, and the source and observer are approaching
each other directly, then the following relationship applies:
f'= f
( v wave + vobserver )
( v wave − v source )
where: f ' is the apparent or observed frequency (in Hz)
f is the original frequency (in Hz)
vwave is the speed of the waves in the medium (in m s−1)
vobserver is the speed of the observer relative to the medium (in m s−1).
It is positive if the observer is moving towards the source and negative if
moving away.
vsource is the speed of the source relative to the medium (in m s−1). It is
positive if the source is moving towards the observer and negative if
moving away.
CHAPTER 9 | SOUND WAVES
255
Worked example 9.2.2
THE DOPPLER EFFECT
A stationary observer hears a car horn as the car drives towards them. The
frequency of the car horn is known to be 300 Hz and the car is driving at
20 m s−1 towards the observer.
What is the observed frequency of the car horn as heard by the stationary
observer? Take the speed of sound to be 340 m s−1.
Thinking
Working
Determine the variables
required to use the
Doppler effect formula.
f = original frequency of sound = 300 Hz
the speed of the waves in the medium
vwave =
= 340 m s−1
vobserver = the speed of the observer relative to the
medium = 0
vsource = the speed of the source relative to the medium
= +20 m s–1 (i.e. moving towards the observer)
Calculate the frequency
for the observer.
f ' = 300
(
340 + 0
340 − 20
)
= 300 × (1.0625)
= 318.75
∴ f ' = 319 Hz.
The observer hears a higher-pitched car horn of
frequency 319 Hz.
Worked example: Try yourself 9.2.2
THE DOPPLER EFFECT
An aeroplane is flying at 150 m s−1 overhead and is emitting a constant sound at
a frequency 1000 Hz.
WS
3.5
What frequency sound would a stationary observer hear once the aeroplane has
passed? Take the speed of sound in air to be 340 m s−1.
Beats
GO TO ➤ Section 8.1, page 234
In Chapter 8, we saw how two (or more) waves can superimpose to either create a
new wave or cancel each other out.
Often in physics we are particularly interested in scenarios where two waves that
are of identical wavelength and amplitude superimpose, either creating a new wave
that is twice the amplitude (constructive interference—when the waves are in phase
with each other) or cancelling each other out (destructive interference—when the
waves are out of phase by half a wavelength.)
For sound waves, superposition creates louder sounds when constructive
interference occurs and softer sounds when destructive interference occurs.
Rather than finding the superposition of waves of different amplitudes, the beats
phenomenon occurs with waves of equal amplitude but different frequencies. The
result (Figure 9.2.6) is a regular pulse-like sound called a beat. The beat frequency
is equal to the difference in the frequencies:
fbeat = f1 − f2
The vertical lines in the beat formula indicate the absolute value or modulus,
which is its non-negative value. For example, |280 − 300| = 20. This is because
(280 − 300) = −20 and the modulus of −20 is 20.
256
MODULE 3 | WAVES AND THERMODYNAMICS
Beats are useful when tuning musical instruments. If the pitch (frequency) of two
instruments is slightly different, a pulsing sound is heard. One musician adjusts the
pitch of their instrument, decreasing the beat frequency until the two instruments
are in tune (i.e. when no beat is heard).
X
Y
Z
f1 =
(a)
f2 =
(b)
f beat =
(c)
FIGURE 9.2.6 Two sound waves with the same amplitude but with slightly different frequencies,
(a) and (b), superimpose to create the pulses or beats shown in (c).
Worked example 9.2.3
BEAT FREQUENCY
Two flutes are tuning up before a performance. They notice a beat frequency of
2 Hz. One of the flute players knows that her flute plays a concert C at exactly
262 Hz.
What frequencies could the other flute be playing?
Thinking
Working
Write out the formula for the beat
frequency.
fbeat = f1 − f2
Substitute the values for the different
frequencies.
fbeat = f1 − f2
2 = |262 − f2|
= |262 − 264| or |262 − 260|
f2 = 264 Hz or 260 Hz.
Worked example: Try yourself 9.2.3
BEAT FREQUENCY
In the physics laboratory, a student is testing two tuning forks. One is labelled
349 Hz, and the other is unlabelled. When struck together, the two tuning forks
produce a beat frequency of 4 Hz.
What are the possible frequencies of the second tuning fork?
CHAPTER 9 | SOUND WAVES
257
9.2 Review
SUMMARY
• Sounds with a greater volume have a larger
amplitude.
• For a mechanical wave, the total Doppler effect
may result from the motion of the source,
the motion of the observer, or the motion of
the medium.
• Sounds that have a higher pitch have a greater
frequency.
(v
• The intensity (I) of a sound wave describes
the rate of energy through an area of space per
second; it is measured in W m−2.
)
• vobserver is positive if the observer is moving towards
the source and negative if moving away.
• The intensity of sound follows an inverse square
relationship:
I∝
+v
f ' = f (vwave − vobserver)
wave source
• vsource is positive if the source is moving towards the
observer and negative if moving away.
1
r2
• If two sound waves of equal amplitude but different
frequencies are superimposed, a pulse or beat is
created. The beat frequency is given by the formula:
• The Doppler effect is a phenomenon that is
observed whenever there is relative movement
between the source of waves and an observer. It
causes an apparent increase in frequency when
the relative movement is towards the observer
and an apparent decrease in frequency when the
relative movement is away from the observer.
fbeat = f1 − f2
KEY QUESTIONS
1
Turning up the volume control on an iPod affects what
sound characteristics?
2
At a concert, sound engineers are worried that the
speakers are producing sound which is above the
pain threshold. They calculate that at 10 m from the
speakers, the intensity is 0.01 W m−2. What is the
intensity of sound 1 m from the speakers? Is this above
the pain threshold?
3
258
A police car travelling at 100 km h−1 along a straight
road has its siren sounding. The police car is pursuing
another car travelling in the same direction, also at
100 km h−1. There is no wind at the time. Would an
observer in the car being pursued hear the siren from
the police car at a higher, lower, or the same frequency
that is emitted by the police car? Explain your answer.
MODULE 3 | WAVES AND THERMODYNAMICS
4
A tourist standing at Circular Quay in Sydney sees
a ferry approaching the dock. The ferry, travelling at
8.0 m s−1, sounds its horn. The horn produces a sound
of frequency 100 Hz. Assuming the speed of sound
over the water is 340 m s−1. Determine the frequency
of the note that the tourist hears.
5
While tuning up for an orchestral rehearsal, three
violins all play a concert A note. The three violins have
slightly different frequencies: 440 Hz, 438 Hz and
441 Hz. If only two violins play at any one time, what
are the possible beat frequencies that these violins
could produce?
9.3 Standing waves
Drawing a bow across a violin string causes the string to vibrate between the fixed
bridge of the violin and the finger of the violinist (Figure 9.3.1). The simplest
vibration has its maximum amplitude at the centre of the string, halfway between
bridge and finger. This is a very simple example of a transverse standing wave.
Standing waves are formed by the superposition of waves. They occur when two
waves of the same amplitude and frequency are travelling in opposite directions
towards each other in the same string. Usually, one wave is the reflection of the
other. Standing waves are responsible for the wide variety of sounds associated with
speech and music.
STANDING WAVES IN A STRING
In Chapter 8, standing waves were explained in terms of continuous wave pulses
travelling along a rope, being reflected at the fixed end and then interfering, causing
a wave that appears to be stationary. It is worth noting that these standing waves
have points where the rope remains still and also points where the rope oscillates
with maximum amplitude.
The points on the rope that remain still, called nodes, are due to destructive
interference. Constructive interference is occurring at the points on the standing
wave where the rope oscillates with maximum amplitude (antinodes).
Nodes and antinodes in a standing wave remain in a fixed position for a particular
frequency of vibration. Figure 9.3.2 illustrates a series of possible standing waves
in a rope, with both ends fixed, corresponding to three different frequencies. The
lowest frequency of vibration (a) produces a standing wave with one antinode in the
centre of the rope. The ends are fixed so they will always be nodal points. Assuming
the tension in the rope remains the same, patterns (b) and (c) are produced at twice
and three times the original frequency respectively.
(a)
(b)
GO TO ➤ Section 8.1, page 237
(c)
antinode
antinode
node
FIGURE 9.3.2
FIGURE 9.3.1 Transverse standing waves can
form along a violin string when the string is
bowed by the violinist.
node
antinode
node
A rope vibrated at three different resonant frequencies, illustrating the standing waves produced at each frequency.
The frequencies at which standing waves are produced are called the resonant
frequencies of the rope. It is important to note that the formation of a standing
wave does not mean that the string or rope itself is stationary. It will continue
to oscillate as further wave pulses travel up and down the rope. It is the relative
positions of the nodes and antinodes that remain unchanged.
It is also important to note that standing waves are not a natural consequence
of every wave reflection. Standing waves are produced only when two waves of
equal amplitude and frequency, travelling in opposite directions, are superimposed.
Standing waves are examples of resonance. They occur only at the natural, or
resonant, frequencies of vibration of the particular medium.
CHAPTER 9 | SOUND WAVES
259
HARMONICS
A large variety of waves is created by the strings of a musical instrument. The
waves travel along the string in both directions and are reflected from the fixed
ends. Most of these vibrations interfere in a random fashion and die away. However,
those corresponding to the resonant frequencies of the string form standing waves
and remain.
The resonant frequencies produced in this complex vibration of multiple standing
waves are termed harmonics. The lowest and simplest form of vibration, with one
antinode (Figure 9.3.2a), is called the fundamental frequency. Higher-level
harmonics (Figure 9.3.2b and c) are referred to by musicians as overtones.
STRINGED INSTRUMENTS
The fundamental frequency usually has the greatest amplitude, so it has the greatest
influence on the sound. The amplitude generally decreases for each subsequent
harmonic. Usually all possible harmonics are produced in a string simultaneously,
and the strings in the instrument and the air around them also vibrate to create the
complex mixture of frequencies that we hear as an instrumental note.
String fixed at both ends
The resonant frequencies or harmonics in a string of length l can be calculated
from the relationship between the length of the string and the wavelength, λ, of the
corresponding standing wave. For a string fixed at both ends (Figure 9.3.3):
• The first harmonic, or fundamental frequency, has one antinode in the centre of
the string and λ = 2 × l.
• The second harmonic has two antinodes and λ = l = 22l .
• The third harmonic has three antinodes, and so on, expressed as λn = 2nl in
general for any harmonic.
λ = 2nl
where: λ is the wavelength (in m)
l is the length of the string (in m)
n is the number of harmonics, which is also the number of antinodes
(1, 2, 3, 4…).
The relationship between wavelength λ and string length l is shown in
Figure 9.3.3.
first harmonic
(fundamental
frequency)
λ1 = 2l
third harmonic
(second overtone)
λ3 =
2l
3
second harmonic
(first overtone)
λ2 = l
fourth harmonic
(third overtone)
λ4 =
l
2
FIGURE 9.3.3 The first four resonant frequencies, or harmonics, in a stretched string fixed at both
ends. Because the ends are fixed, they are always nodal points.
The wave equation v = fλ gives the relationship between frequency, velocity and
string length.
For the first harmonic, or fundamental frequency:
v = fλ and so f = λv = 2vl .
260
MODULE 3 | WAVES AND THERMODYNAMICS
For the second harmonic:
f =
v
λ
= vl
For the third harmonic:
f =
v
λ
= 32vl
Therefore, in general:
f = nv
2l
where: n is the number of the harmonic
f is the frequency of the wave (Hz)
v is the velocity of the wave (m s−1)
l is the length of the string (m)
String fixed at one end
For a string fixed at one end and free at the other, the standing waves that form are
shown in Figure 9.3.4.
first harmonic (fundamental frequency)
N
A
λ1 = 4l
f1 = 4lv
third harmonic (first overtone)
N
A
A
N
λ3 =
4l
3
f3 = 3v
4l = 3f1
fifth harmonic (second overtone)
N
A
N
A
ratio of frequencies f1 : f3 : f5 = 1 : 3 : 5
N
A
λ5 =
4l
5
f5 =
5v
4l
= 5f1
FIGURE 9.3.4 The lower harmonics for a string that is fixed at one end (left-hand side of the diagram)
and free to move at the other (right-hand side of the diagram). Only odd-numbered harmonics are
possible, since only these satisfy the condition of having a node at the fixed end and an antinode at
the free end.
The fixed end of the string is always a node (left-hand side of Figure 9.3.4)
and an antinode always forms at the free end of the string (right-hand side of
Figure 9.3.4).
The first harmonic, or fundamental frequency, has a wavelength four times the
length of the string:
λ=4×l
The next harmonic that can form has a wavelength:
λ=
4× l
3
The next harmonic that can form has a wavelength:
λ=
4× l
5
CHAPTER 9 | SOUND WAVES
261
In general:
PHYSICS IN ACTION
ICT
λ = 4nl
where: λ is the wavelength (in m)
l is the length of the string (in m)
n is the number of the harmonic, odd-number integers only (i.e. 1, 3, 5, …).
Guitar strings
The sound produced by a ukulele
is very different to that of a bass
guitar. This is partly due to the
different length of the instruments,
but the material of the strings also
plays a part.
For stringed instruments, two
other variables affect the frequency
of the sound: the string’s tension
and its mass.
The speed, v, of a wave in a
string can be calculated from
the equation:
v=
T
m/ l
In general, for frequency in terms of velocity:
f = nv
4l
where n is the number of the harmonic (i.e. 1, 3, 5, …).
Notice that only odd-numbered harmonics form, since the conditions necessary
for a standing wave to form are only met when there is an antinode at the free
end and a node at the fixed end. The ratio of the wavelengths of the harmonics is
1 : 3 : 5: …, which means that the wavelength of the third harmonic is 1/3 of the
length of the fundamental frequency (the first harmonic), the fifth harmonic is 1/5
of the length of the fundamental frequency … and so on.
Evenly numbered harmonics (i.e. 2, 4, 6, …) cannot form. The equations
introduced above can be modified to show this. The modified formula is:
where
λ = (2n4−l 1)
T is the tension in the string (in N)
where: λ is the wavelength (in m)
l is the length of the string (in m)
n is the number of antinodes (an integer)
m is the mass (in kg)
l is the length (in m)
This can be included in the
equations for frequency, so that the
frequency of a string fixed at both
ends is given by:
f=
nv
2I
=
n
2I
T
m/ I
As the tension is proportional
to the frequency (f ∝ T ), a string
with more tension produces a
higher-frequency sound.
Similarly, mass is inversely
proportional to frequency
(f ∝
1 ),
m
so heavier strings results
in a lower frequency.
The shorter, low-mass strings of
a ukulele produce a much higher
frequency sound than the long,
heavy strings of a bass guitar.
Note that, for this equation, n is defined as the next harmonic in the sequence,
not the harmonic number as defined for strings fixed at both ends.
It should also be noted that the resonant frequencies of a string depend on
its tension and mass per unit length. Tightening or loosening the string changes
the wavelengths and resonant frequencies for that string (e.g. a musical stringed
instrument is tuned by adjusting the tension in the strings). Heavy strings have
different resonant frequencies than lighter strings of the same length and tension.
Worked example 9.3.1
FUNDAMENTAL FREQUENCY IN STRINGS
A violin string, fixed at both ends, is 22 cm long. It is vibrating at its fundamental
vibration mode of 880 Hz.
a What is the wavelength of the fundamental frequency?
Thinking
Working
Identify the length of the string (l ) in
metres and the harmonic number (n).
n=1
Recall that for any frequency, λ = 2I .
n
Substitute the values from the question
and solve for λ.
262
MODULE 3 | WAVES AND THERMODYNAMICS
l = 22 cm = 0.22 m
λ=
=
2I
n
2×0.22
1
= 0.44 m.
b What is the wavelength of the second harmonic?
Thinking
Working
Identify the length of the string in
metres and the harmonic number.
l = 22 cm = 0.22 m
Recall that for any frequency, λ = .
Substitute the values from the question
and solve for λ.
λ=
n=2
2l
n
2l
n
2 × 0.22
= 2
= 0.22 m
Worked example: Try yourself 9.3.1
FUNDAMENTAL FREQUENCY IN STRINGS
A 25 cm length of string is fixed at both ends. Assume that the tension in the string
is not changed.
a What is the wavelength of the fundamental frequency?
WS
3.7
b What is the wavelength of the third harmonic?
WIND INSTRUMENTS AND AIR COLUMNS
Longitudinal stationary waves are also possible in air columns. These create the
sounds associated with wind instruments. Blowing over the mouthpiece of a flute
(Figure 9.3.5) or the reed of a saxophone produces vibrations that correspond to a
range of frequencies that create sound waves in the tube.
The compressions and rarefactions of the sound waves, confined within the tube,
are reflected from both open and closed ends. This creates the right conditions for
resonance and the formation of standing waves. The length of the pipe determines
the frequency of the resonating sounds.
Open-ended air columns
Instruments such as flutes produce sound in a column or pipe which is open at both
ends. Sound waves are reflected in the pipe from the open ends, so that there are
nodes at both ends of the pipe (Figure 9.3.6).
FIGURE 9.3.5 Blowing over the mouthpiece of a
flute and controlling the length of the flute with
the keys produces a particular note.
first harmonic
(fundamental
frequency)
λ 1 = 2l
v
λ1
f1 =
v
2l
=
second harmonic
(first overtone)
λ2 = l
f2 =
v
λ2
=
v
l
= 2f1
third harmonic
(second overtone)
λ3 =
2l
3
λ4 =
l
2
f3 =
v
λ3
=
3v
2l
= 3f1
fourth harmonic
(third overtone)
FIGURE 9.3.6
f4 =
v
λ4
=
2v
l
= 4f1
The first harmonics for a pipe that is open at both ends.
CHAPTER 9 | SOUND WAVES
263
For a pipe open at both ends:
λ = 2nl
where: λ is the wavelength (in m)
l is the length of the pipe (in m)
n is the number of the harmonic
In general, for frequency in terms of velocity:
f = nv
2l
Worked example 9.3.2
HARMONIC WAVELENGTHS IN OPEN-ENDED AIR COLUMNS
An open-ended pipe 62 cm long has sound travelling in it at 300 m s−1.
a What is the wavelength of the third harmonic?
Thinking
Working
Identify the length of the air column (l )
in metres and the harmonic number (n).
n=3
Recall that for any wavelength, λ =
λ=
2l
n
.
Substitute the values from the question
and solve for λ.
l = 62 cm = 0.62 m
=
2l
n
2×0.62
3
= 0.41 m
b What is the frequency of the third harmonic?
Thinking
Working
Identify the wavelength (λ) of the note in
metres and the speed of sound.
λ = 0.41 m from (a) above
Use the wave equation v = f λ .
f=
v = 300 m s−1
v
λ
=
300
0.41
= 731.7
= 732 Hz
Worked example: Try yourself 9.3.2
HARMONIC WAVELENGTHS IN OPEN-ENDED AIR COLUMNS
A small flute (a piccolo) has a length of 32 cm and the sound has a speed of
320 m s−1. Consider the piccolo to be an open-ended pipe.
a What is the wavelength of the fourth harmonic?
b What is the frequency of the fourth harmonic?
264
MODULE 3 | WAVES AND THERMODYNAMICS
Closed air columns
In a closed air column, the pipe is closed at one end. The open end of the pipe
reflects the sound, creating a phase change and a pressure node. At the closed end,
constructive interference occurs and an antinode is formed. This means there is a
node at one end and an antinode at the other (Figure 9.3.7).
In general, only odd-numbered harmonics exist for pipes open at one end.
λ = 4nl
f = nv
4l
where: λ is the wavelength (in m)
l is the length of the pipe (in m)
n is the number of the harmonic; n = 1, 3, 5, … (odd numbers only).
l
first harmonic
(fundamental frequency)
λ 1 = 4l
N
A
f1 = 4lv
third harmonic
(first overtone)
A
N
A
N
λ 3 = 4l3
f3 =
3v
4l
= 3f1
fifth harmonic
(second overtone)
N
A
A
N
A
N
λ 5 = 4l5
f5 =
5v
4l
= 5f1
ratio of frequencies f1 : f3 : f5 = 1 : 3 : 5
FIGURE 9.3.7 In a pipe closed at one end there is a node (N, left) at one end and an antinode
(A, right) at the other end. This means that only odd-numbered harmonics exist.
Worked example 9.3.3
FUNDAMENTAL FREQUENCY IN CLOSED AIR COLUMNS
A clarinet functions like a pipe closed at one end. Sound in it travels at a velocity of
300 m s−1. It has a fundamental frequency of 125 Hz.
a What is the length of the clarinet? Give your answer in centimetres.
Thinking
Working
−1
Identify the frequency, f (in Hz), the speed, v (in m s )
and the harmonic number, n.
f = 125 Hz
n=1
v = 300 m s−1
Recall that for any wavelength, λ =
4l .
n
Substitute the values from the question and solve for λ.
f=
125 =
nv l
4l
1 × 300
4l
l = 0.6 m = 60 cm
CHAPTER 9 | SOUND WAVES
265
b What is the wavelength of the next possible harmonic above the fundamental
frequency that can be produced in this clarinet?
Thinking
Identify the length of the pipe (l) in metres and the
harmonic number (n).
Recall the formula for any wavelength.
Substitute the values from the question and
solve for λ.
Working
l = 0.6 m
n = 3 because only odd
harmonics are possible
for a closed-end pipe
λ=
=
4l
n
4×0.6
3
= 0.8m
Worked example: Try yourself 9.3.3
FUNDAMENTAL FREQUENCY IN CLOSED AIR COLUMNS
An organ pipe can produce a fundamental frequency of 126 Hz. The speed of
sound in the pipe is 340 m s−1. Organ pipes are closed at one end.
a What is the length of the pipe? Give your answer in centimetres.
PA
3.2
PA
3.2
b What is the wavelength of the next possible harmonic that can be produced in
this pipe?
PHYSICSFILE ICT
The didgeridoo
Just as with strings that are fixed
at one end and open at the other,
the formula for the wavelength
can be rearraged to:
λ = (2n4−l 1)
where n is now the number of
antinodes.
266
The didgeridoo is made from a tree branch or trunk, traditionally about 1.4 m long
and with an irregular internal diameter. The mouthpiece end is coated with beeswax,
allowing the player to seal that end and also to make it more comfortable for the player.
Like a trumpet or a trombone, the player’s lips vibrate, producing the driving frequency.
If this matches the natural frequency of the didgeridoo, it resonates and produces
large-amplitude sounds.
Because the didgeridoo has a fixed length, it produces only a single fundamental
frequency. For a didgeridoo 1.4 m long, the fundamental frequency is approximately
60 Hz (from f1 = v/4l). Because it is closed at one end, it produces only
odd-numbered harmonics.
Experienced didgeridoo players make particular frequencies more dominant by
adjusting the frequency of the vibrations of their lips. They can also produce other
sounds using their vocal cords and by using the space in their own mouth as a
resonating chamber. The continuous sound of a didgeridoo is produced by a process
known as circular breathing. To keep the lips vibrating, air must be continuously blown
out through the lips. A constant stream of air is achieved by inhaling and storing air in
the mouth and squeezing it through the lips using the cheek muscles, at the same time
breathing in through the nose. This is a difficult skill to master!
At a simple level, the didgeridoo may be considered to be a closed-end pipe (closed
at one end by the lips of the player), but studies of the physics of the didgeridoo have
shown that it is more complex than first thought.
MODULE 3 | WAVES AND THERMODYNAMICS
9.3 Review
SUMMARY
• For a string fixed at one end or a pipe closed at
one end, the wavelength of the standing waves
corresponding to the various harmonics is:
• Standing or stationary waves occur as a result of
resonance at the natural frequency of vibration.
• Points on a standing wave that remain still are
called nodes.
• Points of maximum amplitude on a standing wave
are called antinodes.
f=
nv
2l
f=
nv
4l
Only odd-numbered harmonics may be present.
• The speed of sound is dependent on the medium.
• Within a string fixed at both ends or a pipe open
at both ends, the wavelength of the standing
waves corresponding to the various harmonics is:
2l
n
4l
n
and the frequency is:
• The standing wave frequencies are referred to as
harmonics. The simplest mode is referred to as
the fundamental frequency.
λ=
λ=
• The tension and mass of the string affect the
speed of sound by the following relationship:
v=
T
m/ l
and the frequency is:
All harmonics may be present.
KEY QUESTIONS
1
What is the wavelength of the fundamental mode of
a standing wave on a string 0.4 m long and fixed at
both ends?
2
Calculate the length of a string fixed at both ends
when the wavelength of the fourth harmonic is 0.75 m.
3
A standing wave is produced in a rope fixed at
both ends by vibrating the rope with four times
the frequency that produces the fundamental or
first harmonic. How much larger or smaller is the
wavelength of this standing wave compared to that of
the fundamental or first harmonic?
4
The fundamental frequency of a violin string is 350 Hz
and the velocity of the waves along it is 387 m s−1.
What is the wavelength of the new fundamental mode
when a finger is pressed to shorten the string to
two-thirds of its original length?
5
A metal string 50 cm long (at constant tension) is
plucked, creating a wave pulse. The speed of the
transverse wave created is 300 m s−1. Both ends of the
string are fixed.
a Calculate the fundamental frequency.
b Calculate the frequency of the second harmonic.
c Calculate the frequency of the third harmonic.
6
A flute can be considered to be an open-ended air
column. For a flute 50 cm long, in which the speed of
sound is 320 m s−1:
a What is the wavelength of the second harmonic
produced by the flute?
b What is the wavelength of the third harmonic
produced by the flute?
c Calculate the frequency of the third harmonic
produced by the flute.
7
A pipe 70 cm long is open at one end and closed at the
other. The speed of sound inside the pipe is 315 m s−1.
a What is the fundamental frequency of the pipe?
b What is the frequency of the fifth harmonic?
CHAPTER 9 | SOUND WAVES
267
Chapter review
KEY TERMS
frequency
fundamental frequency
harmonic
infrasonic
intensity
inverse square law
absolute value
antinode
beat
beat frequency
compression
Doppler effect
node
overtone
rarefaction
standing wave
ultrasonic
REVIEW QUESTIONS
The following information applies to questions 1 to 4.
The digital output from two different sound waves
(A and B) is shown in the figure below. The graphs describe
the changes in air pressure over time.
B
Particle displacement
A
The following information relates to questions 6 to 8.
The diagram shows the displacement of the air molecules
in a sound wave from their mean positions as a function of
distance from the source, at a particular time. The wave is
travelling to the right at 340 m s−1.
1
2
What is the ratio of the amplitudes A:B?
A 1:1
B 5:8
C 8:5
D 1:4
What is the ratio of the frequencies A:B?
A 1:1
B 2:1
C 1:2
D 1:4
3
Which of the two sound graphs depicts a higher pitch?
Justify your answer.
4
Which of the two sounds has a higher volume? Justify
your answer.
5
Sound waves are longitudinal waves with areas of
higher and lower pressure. Which one (or more) of the
following statements is true?
A Pressure variation is at a maximum at a rarefaction.
B Pressure variation is at a minimum at a rarefaction.
C Pressure variation is zero between a compression
and a rarefaction.
D Pressure variation is zero at a rarefaction and a
maximum at a compression.
268
MODULE 3 | WAVES AND THERMODYNAMICS
1
0
–1
1.0
2.0
3.0
4.0
Y
5.0
6.0
Distance from source (m)
6
What is the amplitude of the sound wave?
A 1.0
B 2.0
C 3.0
D 4.0
7
What is the wavelength of the sound wave?
A 2.0 m
B 3.0 m
C 4.0 m
D 5.0 m
8
In
A
B
C
D
9
Which one or more of the following sound properties is
dependent on the source producing the sound?
A intensity
B frequency
C speed
D volume
what direction is a particle at point Y moving?
up
down
left
right
10 A motorbike travelling at constant speed produces
a long, steady sound. You are unable to see the
motorbike, but you hear its sound suddenly change
pitch. Which one or more of the following options best
explains the motion of the motorbike relative to you?
A The motorbike was always travelling towards you.
B The motorbike was always travelling away from you.
C The motorbike travelled away from you, then
towards you.
D The motorbike travelled towards you, then away
from you.
11 At a Formula One motor car race, a race car is travelling
down a straight section of road towards you at a speed
of 180 km h−1. The engine noise may be considered to
have a constant frequency of 2000 Hz. The speed of
sound is 340 m s−1. If you remain stationary:
a What frequency does the engine sound have as the
car approaches you?
b What frequency does the engine sound have as the
car travels away from you?
12 In a medium where a mechanical wave can be
propagated, when is the Doppler effect observable?
13 A loudspeaker is emitting a sound with an intensity of
2.2 × 10−5 W m−2 at a distance of 3 m from the speaker.
Calculate the intensity at a distance of 8 m.
14 A signal generator produces vibrations in a string that
is fixed at one end and free to move at the other. The
effective length of the string is 85 cm. The speed of the
vibrations along the string is 340 m s−1.
a What is the lowest frequency of vibration that will
produce a standing wave in the string?
b What is the frequency of the third harmonic?
15 An earthquake causes a footbridge to oscillate up and
down with a fundamental frequency once every 4.0 s.
The motion of the footbridge can be considered to
be like that of a string fixed at both ends. What is the
frequency of the second harmonic for this footbridge?
16 The velocity of waves in a particular string at constant
tension is 78 m s−1. The string is fixed at both ends.
If the frequency of a particular standing wave in the
string is 428 Hz, how far apart would two adjacent
antinodes be?
17 Sound is produced in pipe that is 64 cm long and
closed at one end.
a What is the fundamental frequency of the pipe if the
speed of the sound is 320 m s−1?
b What is the wavelength of the fundamental
frequency?
c What frequency of sound will cause the pipe to
resonate at its third harmonic?
18 Explain why the strings on a guitar have different
thicknesses.
19 You are conducting an investigation into how people
produce sound from their vocal cords. You create a
model which compares the vocal cords to a string
fixed at both ends. Assuming this is a reasonable
approximation, calculate the length of the cords needed
to produce a fundamental frequency of 200 Hz. Take
the speed of sound in the vocal cords to be 8 m s−1.
20 After completing the activity on page 246, reflect on the
inquiry question: What evidence suggests that sound
is a mechanical wave? In your response, discuss how
the change in sound intensity in the activity can explain
that sound is a wave.
CHAPTER 9 | SOUND WAVES
269
CHAPTER
Ray model of light
Discovering the nature of light has been one of the scientific community’s greatest
challenges. Over the course of history, light has been compared to a geometric ray,
a stream of particles or even a series of waves. However, these relatively simple
models have been found to be limited in their ability to explain all of the properties
of light.
The search for a more adequate model has pushed scientists to develop new types
of equipment and more sophisticated experiments. Over time, it has also led to a
reshaping of the understanding of the nature of matter and energy.
This chapter explores the ray model of light—a simple yet powerful way of
describing many phenomena associated with light.
Content
INQUIRY QUESTION
What properties can be demonstrated when using the ray
model of light?
By the end of this chapter you will be able to:
• conduct a practical investigation to analyse the formation of images in mirrors
and lenses via reflection and refraction using the ray model of light (ACSPH075)
• conduct investigations to examine qualitatively and quantitatively the refraction
and total internal reflection of light (ACSPH075, ACSPH076)
• predict quantitatively, using Snell’s law, the refraction and total internal reflection
of light in a variety of situations CCT
• conduct a practical investigation to demonstrate and explain the phenomenon of
the dispersion of light CCT
• conduct an investigation to demonstrate the relationship between the inverse
square law, the intensity of light and the transfer of energy (ACSPH077)
• solve problems or make quantitative predictions in a variety of situations by
applying the following relationships: ICT N
-- nx = vcx for the refractive index of medium x, where vx is the speed of light in
the medium
-- n1 sin i = n2 sin r (Snell’s law)
-- sin ic =
1
nx
for the critical angle ic of medium x
2
2
-- I1r1 = I2r2 to compare the intensity of light at two points, r1 and r2.
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
10.1 Light as a ray
The ancient Greeks were among the first people to develop a model to explain
the behaviour of light. Drawing on the work of mathematicians such as Euclid,
they imagined that light behaved like a series of rays radiating outwards from a
single source and being detected by our eyes. This fits with the way light can often
be perceived in a dark and dusty room (Figure 10.1.1). Over the centuries since,
scientists have replaced the relatively simple ray idea with a variety of much more
sophisticated models for light. However, the ray model is a useful starting point for
the study of light.
FIGURE 10.1.1
Light can be considered to be a series of rays radiating outwards from a single point.
THE LAW OF REFLECTION
When a light ray strikes a solid object, the object absorbs some of the light and some
of the light may be transmitted through the object (if it is transparent or translucent).
Any remaining light is reflected from the surface of the object. In nature, we observe
two basically different types of reflection: regular reflection and diffuse reflection.
Regular reflection occurs from a very even surface. Parallel rays of light falling on
the surface remain parallel when reflected (Figure 10.1.2a). If the surface is rough
or uneven, then diffuse reflection occurs. Parallel rays of light falling on an uneven
surface are reflected diffusely (Figure 10.1.2b).
(a)
(b)
FIGURE 10.1.2 (a) Regular reflection from a smooth surface occurs when parallel rays of incident
light are reflected parallel to each other. (b) Diffuse reflection occurs at an irregular surface. Here, the
incoming parallel rays are reflected at all angles.
It is possible to predict the direction of a reflected light ray using the law of
reflection. Consider a single light ray directed onto a plane mirror as shown in
Figure 10.1.3. A normal is constructed at the point where the ray strikes the mirror.
272
MODULE 3 | WAVES AND THERMODYNAMICS
The ray of light that strikes the mirror is called the incident ray; the ray of light that
is reflected from the mirror is called the reflected ray. The angle of incidence (i ) and
the angle of reflection (r) are defined with respect to the normal.
The law or reflection states that
the angle of incidence is equal to
the angle of reflection, i.e. i = r.
plane (flat) mirror
inc
ide
normal
FIGURE 10.1.3
reflection: i = r.
nt
ray
angle of incidence
i
angle of reflection r
ray
ed
ect
fl
re
When light reflects from a plane mirror, the angle of incidence equals the angle of
PLANE MIRRORS
Although we may be consciously aware that light is reflected from a mirror, when
our brain processes visual information, it assumes that light travels in a straight line.
This means that when we stand in front of a plane mirror, we perceive an image of
ourselves behind the mirror.
One way to explain why this happens is to use a ray diagram. As its name
suggests, this is a diagram that represents light as straight rays.
In Figure 10.1.4a, the ray diagram shows two rays of light travelling outwards
from Reuben’s foot. When using the ray model of light we would assume that there
is actually an infinite number of rays heading out in all directions from his toe.
However, for the purposes of a ray diagram, it’s only necessary to draw the rays of
light that would reach the eyes of the observer. (Note: the rays from Reuben’s foot
are actually reflected rays that would have originally come from some light source
like the Sun or a light bulb.)
The light rays are represented as straight lines with arrows indicating the direction
in which the light is travelling. One of the rays is seen by each of Jessie’s eyes. Jessie’s
brain uses a process of triangulation to judge the position of Reuben’s toe. Similar
diagrams could be drawn to show how Jessie can see other parts of Reuben’s body.
Figure 10.1.4b shows what happens when Jessie stands in front of a large plane
mirror. We can once again assume that rays are coming out in all directions from
Jessie’s foot. In the ray diagram we have shown the two rays that would hit the
mirror and be reflected back into Jessie’s eyes. Jessie’s brain assumes that each ray
travels in a straight line, as shown by the virtual rays. Therefore she sees an image of
herself standing on the opposite side of the mirror.
This type of image seen in a plane mirror is known as a virtual image. A virtual
image is one that is formed where light rays appear to cross; it can only be seen by
looking into the mirror. As we will see later, some mirrors can form real images
where the light rays actually cross. Real images can only be observed by projecting
them onto a screen.
We describe the image formed by a plane as upright (meaning that it appears in
the same orientation) and the same size as the object that formed it. The image in
a plane mirror also appears to be the same distance from the mirror as the object
that formed it.
When describing the image formed by a mirror, we usually refer to its:
• type (real or virtual)
• orientation (upright or inverted)
• magnification (diminished, enlarged or true size)
• position (the distance between the image and the mirror).
(a)
A
Jessie
Reuben
2m
(b)
real rays
virtual rays
Jessie
1m
mirror
Jessie’s reflection
1m
FIGURE 10.1.4 (a) Diverging rays from Reuben’s
foot are focused by Jessie’s eye. (b) Diverging
reflected rays enter Jessie’s eye and appear
to have come from the foot of a girl standing
2 m away.
CHAPTER 10 | RAY MODEL OF LIGHT
273
PHYSICSFILE ICT
Lateral inversion
The image that you see in a mirror is not exactly what you look like. The geometry of
reflection means that images are reversed from left to right—a process known as lateral
inversion. If you stand in front of a mirror and hold up your right hand, it will appear as
if your image is holding up its left hand (Figure 10.1.5).
FIGURE 10.1.5 Although the woman is applying makeup with her right hand, in her image she
appears to be applying makeup with her left hand.
Unlike a mirror image, a photo of your face is not laterally inverted, so it will look
slightly different to the face that you are used to seeing in the mirror. However, the
differences may not be so great that you consciously notice them. Instead, you might
just feel like something is ‘wrong’ with the photo.
INTENSITY
FIGURE 10.1.6 The unit of luminous intensity is
based on a standard candle.
274
In mathematics, a ray continues on infinitely in one direction. In reality, a light
ray does not continue infinitely; it decreases in intensity the further it gets from
its source.
In physics, intensity is a measure of the rate at which energy passes through an
area of space. This corresponds roughly to what we would call the ‘brightness’ of
the light.
The SI unit of luminous intensity is the candela (cd). As its name suggests, this
unit was originally based on the construction of a candle of standard brightness
(Figure 10.1.6). A typical household candle produces approximately 1 cd.
Consider Figure 10.1.7. As light radiates outwards from a point source, the rays
spread further and further apart.
Let us assume that the light energy flowing through the shaded box at a distance
r represents an intensity of I1. At 2r, the same amount of light energy is passing
through an area four times larger than at r. This means that the light at this point will
be only one-quarter as bright as at r, i.e. I 2 = 41 I1 . Similarly, at 3r, the light energy is
flowing through an area nine times bigger than at r, hence I 3 = 91 I1 .
From this, we can see that the intensity of the light is inversely proportional to
the square of the distance from the light source, i.e. I ∝ 12 .
MODULE 3 | WAVES AND THERMODYNAMICS
r
The relationship between the intensity of light at two points at different
distances from a light source is given by:
3r
I1r12 = I2 r22
2r
where:
I1 and I2 are the intensities at points 1 and 2 respectively (in cd)
r1 and r2 are the distances of points 1 and 2 respectively from the light
source (in m).
light
source
r
Worked example 10.1.1
CALCULATING INTENSITY
The light from a 25 W compact fluorescent light bulb has an intensity of 11 cd at
a distance of 1 m from the bulb.
Calculate the intensity of the light at a distance of 2 m from the bulb.
Thinking
Working
Recall the formula for intensity.
I1r12 = I2r22
Rearrange the formula to make I2 the
subject.
I2 =
Substitute the appropriate values into
the formula and solve.
I2 = 11×21 = 2.75 cd
I1r12
r22
FIGURE 10.1.7 As it moves away from the light
source, the light spreads out radially. As the
radius increases, the same amount of energy
passes through a larger area. The area increases
with the square of the radius, so the intensity
decreases by this ratio.
PHYSICSFILE N
Inverse square laws
2
2
Worked example: Try yourself 10.1.1
CALCULATING INTENSITY
The intensity of the light 50 cm away from an LED (i.e. light-emitting diode) is
0.016 cd.
In physics, there are a number of
phenomena that decrease in strength
or intensity as the square of distance
increases. In Chapter 9 we saw that
the intensity of sound waves varies
according to an inverse square
relationship. Gravitational, electrostatic
and magnetic effects all follow this
same type of relationship.
Calculate the intensity of the light at a distance of 10 cm from the LED.
Worked example 10.1.2
PHYSICSFILE ICT
CALCULATING INTENSITY CHANGES USING PROPORTIONAL REASONING
Luminous intensity
The distance between a compact fluorescent light bulb and the observer is doubled.
Using the SkillBuilder on page 276, how much will the intensity of the light from
the bulb change?
Thinking
Working
Recall the formula for intensity.
I1r12 = I2r22
Describe the change to the radius as
an equation.
r2 = 2r1
Substitute this relationship into the
formula for intensity.
I1r12 = I2 (2r1 )2
Simplify this equation to describe the
change in intensity.
I1r12 = I2 × 4r12
∴ I1 = I2 × 4
∴ I2 =
Interpret the answer.
The human eye is more sensitive to
some colours than others. For example,
our eyes are most sensitive to greenishyellow light. There are also types of
light in the light spectrum, like infra-red
or ultra-violet light, which our eyes
cannot see at all.
When discussing our perception of the
brightness of a light source, scientists
use complex mathematical models
which compensate for our eyes’ colour
sensitivity.
1I
4 1
The intensity is reduced to a quarter.
CHAPTER 10 | RAY MODEL OF LIGHT
275
Worked example: Try yourself 10.1.2
PA
3.5
DS
3.1
PA
3.5
CALCULATING INTENSITY CHANGES USING PROPORTIONAL REASONING
Calculate the change in the intensity of the light from an LED (light-emitting
diode) if it is brought 5 times closer.
SKILLBUILDER
N
Using proportional reasoning in Physics
If we know the type of relationship between two variables
(e.g. direct, quadratic, inverse, inverse square), we can use
our knowledge of proportion to determine how much one
variable will change due to a change in the other variable.
Consider a simple situation where there is a direct
relationship between the two variables x and y (i.e. x ∝ y).
If we are told that the value of x will double, then we
can predict that y will also double. This is an example of
‘proportional reasoning’.
Similarly, if there is an inverse relationship between x
and y (i.e. x ∝ 1y ), then we know if the value of x is doubled,
the value of y will be halved.
There are many different strategies that can be used to
solve problems involving proportional reasoning. One
reliable algorithm is outlined below.
1 Describe the relationship between variables as a
statement of proportion, e.g. x ∝ 1y .
2 Rewrite the proportionality as an equation by
introducing a constant of proportionality, e.g. x = ky .
276
MODULE 3 | WAVES AND THERMODYNAMICS
3 Transpose the equation to make the constant the
subject, e.g. k = xy.
4 Use the fact that the constant of proportionality does
not change for any relationship between the original set
of values of variables (e.g. x1 and y1) and the new set of
values (e.g. x2 and y2). That is, k = x 1y 1 = x 2 y 2.
5 Describe the change to one variable as an equation,
e.g. if x1 is doubled, then x 2 = 2x 1.
6 Substitute the equation from step 5 into the equation
from step 4, i.e. x1y1 = ( 2x1 ) y2.
7 Simplify this new equation to describe the change in the
other variable by cancelling x 1 from both sides of the
equation; then y1 = 2 y2 ∴ y2 = 1 y1.
2
8 Interpret the answer: in this case, one interpretation is
that y has been halved.
Note: This process is made much simpler when physical
relationships are expressed in such a way that steps 1 to 4
have already been completed, e.g. I1r12 = I2r22.
10.1 Review
SUMMARY
• Light can be represented as a ray.
• Luminous intensity is a measure of brightness or
the rate at which light energy passes through an
area of space. The SI unit for luminous intensity is
the candela (cd).
• The law of reflection states that the angle of
incidence is equal to the angle of reflection,
i.e. i = r.
• The relationship between the intensity of light
at two points at different distances from a light
source is given by I1r12 = I2r22.
• Ray diagrams can be used to explain the images
seen in plane mirrors.
KEY QUESTIONS
1
Complete the ray paths in the following diagram to
locate the image of the object in the plane mirror.
4
Explain why you cannot see an image of your face
reflected off the page of the book in front of you when
you are reading.
5
If the distance between a light source and an observer
is increased by a factor of 2.5, the intensity of the light
will:
A decrease by a factor of 1.25
B increase by a factor of 1.5
C increase by a factor of 2.5
D decrease by a factor of 6.25
6
For a person sitting 75 cm away from a computer
monitor, the intensity of the light from the monitor is
6.0 cd. What would be the intensity if they moved to sit
only 50 cm from the monitor?
observer’s eye
object
mirror
2
Is
a
b
c
3
A girl stands 7 m away from a full-length vertical plane
mirror.
a How far must she walk to appear to be 3 m from
her image?
b She walks towards the mirror at 1.5 m s–1. At what
speed does she appear to approach her image?
the image formed in a plane mirror always:
upright or inverted?
enlarged, diminished or the same size as the object?
real or virtual?
CHAPTER 10 | RAY MODEL OF LIGHT
277
PHYSICS INQUIRY
CCT
Ray model of light
What properties can
be demonstrated when
using the ray model of
light?
10.2 Refraction
When light passes from one medium (substance) into another, it either speeds up or
slows down. This change in speed causes the light ray to change direction, as shown
in Figure 10.2.1. Refraction is the name given to a change in the direction of light
caused by changes in its speed (Figure 10.2.1).
COLLECT THIS…
•
a large beaker
•
laser light
•
clear water beads
•
white paper
DO THIS…
1 Soak the water beads in water.
2 Place the beaker on the white
paper and half fill with water.
3 Shine a laser light through the
water at an angle. On the paper,
mark the path into and out of
the beaker.
4 Place the water beads into the
beaker. Adjust the water so that
half of the beads are covered
in water and half are out of
the water.
5 Using the laser pointer, shine
through the bottom of the
beaker at the same angle
initially marked on the paper.
Mark the path of the light on
the paper in a different colour.
6 Using the laser pointer, shine
through the top of the beaker.
How does the path the light
takes differ?
FIGURE 10.2.1 Light refracts as it moves from one medium (i.e. the semicircular glass prism) into
another (i.e. air) causing a change in direction.
REFRACTIVE INDEX
The amount of refraction that occurs depends on how much the speed of light
changes as light moves from one medium to another—when light slows down
greatly, it will undergo significant refraction.
The speed of light in a number of different materials is shown in Table 10.2.1.
TABLE 10.2.1
The speed of light in various materials correct to three significant figures.
Material
Speed of light (× 108 m s−1)
vacuum
3.00
air
3.00
ice
2.29
Present your results describing the
path of the light.
water
2.25
quartz
2.05
REFLECT ON THIS…
crown glass
1.97
What properties can be
demonstrated when using the ray
model of light?
flint glass
1.85
diamond
1.24
RECORD THIS…
Describe how we see clear objects.
When you look at a glass, how can
you see if the glass is full of air,
water or another clear substance?
278
MODULE 3 | WAVES AND THERMODYNAMICS
Scientists find it convenient to describe the change in speed of a wave using a
property called the refractive index. The refractive index of medium x, nx, is defined
as the ratio of the speed of light in a vacuum, c, to the speed of light in the medium, vx:
nx =
c
vx
,
where:
nx is the refractive index of medium x
c is the speed of light in a vacuum
vx is the speed of light in the medium.
Note that nx is dimensionless, i.e. it has no units, it is just a number.
The refractive index for various materials is given in Table 10.2.2.
This quantity is also sometimes referred to as the ‘absolute’ refractive index
of the material, to distinguish it from the ‘relative’ refractive index that might
be used when a light ray moves from one medium to another, e.g. from water
to glass.
PHYSICSFILE L
Optical density
For convenience, physicists sometimes
use the term ‘optical density’ to refer
to the differences in refractive index
of materials. A medium with a higher
refractive index is said to be more
‘optically dense’. It is important not
to confuse this term with the physical
density of the material (i.e. its mass/
volume ratio). A medium that is more
physically dense will not necessarily be
more optically dense.
TABLE 10.2.2
materials.
Refractive indexes of various
Material
Refractive index, nx
PHYSICSFILE IU
vacuum
1.00
Debates about the speed of light
air
1.00
In the late 17th century a debate raged among scientists about the speed of light.
The famous English scientist Sir Isaac Newton explained light in terms of particles or
‘corpuscles’, with each different colour of the spectrum representing a different type of
particle. Scientists Robert Hooke (from England) and Christiaan Huygens (from Holland)
proposed an alternative model that described light as a type of wave, similar to the
water waves observed in the ocean.
A key point of difference between the two theories was that Newton’s ‘corpuscular’
theory suggested that light would speed up as it travelled through a solid material such
as glass. In comparison, the wave theory predicted that light would be slower in glass
than in air.
Unfortunately, at that time it was impossible to measure the speed of light accurately,
so the question could not be resolved scientifically. Newton’s esteemed reputation
meant that for many years his corpuscular theory was considered correct.
It was not until the early 19th century that experiments first convincingly demonstrated
that light slows down when it moves from air into glass.
Today, a modern understanding of light draws on aspects of both theories and
is, perhaps, more complex than either Newton, Hooke or Huygens could ever
have imagined.
ice
1.31
water
1.33
quartz
1.46
crown glass
1.52
flint glass
1.62
diamond
2.42
Worked example 10.2.1
CALCULATING REFRACTIVE INDEX
The speed of light is 2.25 × 108 m s−1 in water and 3.00 × 108 m s−1 in a vacuum.
Calculate the refractive index of water.
Thinking
Working
Recall the definition of refractive index.
nwater =
Substitute the appropriate values into
the formula and solve.
nwater =
c
v water
3.00 × 108
2.25 × 108
=
3.00
2.25
= 1.33
CHAPTER 10 | RAY MODEL OF LIGHT
279
Worked example: Try yourself 10.2.1
CALCULATING REFRACTIVE INDEX
The speed of light in crown glass (a type of glass used in optics) is 1.97 × 108 m s−1.
Given that the speed of light in a vacuum is 3.00 × 108 m s−1, calculate the
refractive index of crown glass.
By definition, the refractive index of a vacuum is exactly 1, since nvacuum = cc = 1.
Similarly, the refractive index of air is effectively equal to 1, because the speed of
light in air is practically the same as its speed in a vacuum.
The definition of refractive index allows you to determine changes in the speed
of light as it moves from one medium to another (Figure 10.2.2).
nx = vc , so c = nxvx.
x
This applies for any material. Therefore:
n1v1 = n2v2
where:
n1 is the refractive index of the first material
v1 is the speed of light in the first material
n2 is the refractive index of the second material
v2 is the speed of light in the second material.
Worked example 10.2.2
CHANGES TO THE SPEED OF LIGHT
FIGURE 10.2.2 Light is refracted multiple times
as it travels from air into glass, through the
water and then back out into glass and air.
incident ray
A ray of light travels from crown glass (n = 1.52), where it has a speed of
1.97 × 108 m s−1, into water (n = 1.33).
Calculate the speed of light in water.
Thinking
Working
Recall the formula.
n1v1 = n2v2
Substitute the appropriate values into
the formula and solve.
1.52 × 1.97 × 108 = 1.33 × v2
Worked example: Try yourself 10.2.2
n1
air
water
n2
r
8
= 2.25 × 108 m s−1
normal
i
1.97 × 10
∴ v2 = 1.52 × 1.33
CHANGES TO THE SPEED OF LIGHT
A ray of light travels from water (n = 1.33), where it has a speed of
2.25 × 108 m s–1, into flint glass (n = 1.85).
Calculate the speed of light in flint glass.
SNELL’S LAW
refracted ray
FIGURE 10.2.3 Light refracts towards the normal
as it moves from air into water.
280
Refractive indexes can also be used to determine how much a light ray will refract as it
moves from one medium to another. Consider the situation shown in Figure 10.2.3,
where light refracts as it moves from air into water. Similar to the law of reflection, the
ray coming in from the air is called the incident ray and the ray in the water is called
the refracted ray; the angles of incidence and refraction are defined with respect to a
normal constructed at right-angles to the interface between air and water.
MODULE 3 | WAVES AND THERMODYNAMICS
In 1621, the Dutch mathematician Willebrord Snell described the geometry of
this situation with a formula now known as Snell’s law.
n1 sin i = n2 sin r
where:
n1 is the refractive index of the incident medium
n2 is the refractive index of the refracting medium
i is the angle of incidence
r is the angle of refraction.
Worked example 10.2.3
(a)
normal
r
air
n2
water
n1
USING SNELL’S LAW
A ray of light in air strikes the surface of a pool of water (n = 1.33) at an angle of
30° to the normal.
i
Calculate the angle of refraction of the light in water.
(b)
Thinking
Working
Recall Snell’s law.
n1 sin i = n2 sin r
Recall the refractive index of air.
n1 = 1.00
Substitute the appropriate values into
the formula to find a value for sin r.
1.00 × sin 30° = 1.33 × sin r
normal
r
air
n1
water
× sin 30°
∴ sin r = 1.00 1.33
i
= 0.3759
Calculate the angle of refraction.
n2
r = sin–1 0.3759 = 22.1°
Worked example: Try yourself 10.2.3
USING SNELL’S LAW
(c)
normal
90°
air
n2
n1
water
A ray of light in air strikes a piece of flint glass (n = 1.62) at 50° to the normal.
Calculate the angle of refraction of the light in the glass.
i = ic
TOTAL INTERNAL REFLECTION
When light passes from a medium with low refractive index to one with a higher
refractive index (like the air-to-water example shown in Figure 10.2.3), it is
refracted towards the normal. Conversely, when light passes from a medium with
a high refractive index to one with a lower refractive index, for instance from water
into air as in Figure 10.2.4, it is refracted away from the normal (Figure 10.2.4a).
In this case, as the angle of incidence increases, the angle of refraction gets closer
to 90° (Figure 10.2.4b).
Eventually, at an angle of incidence known as the critical angle, the angle of
refraction becomes 90° and the light is refracted along the interface between the two
mediums (Figure 10.2.4c). If the angle of incidence is increased above this value,
the light ray does not undergo refraction; instead it is reflected back into the original
medium, as though striking a perfect mirror (Figure 10.2.4d). This phenomenon is
known as total internal reflection and is seen in action in Figure 10.2.5.
The angle of refraction for the critical angle is 90°, so the critical angle is defined
by the formula n1 sin ic = n2 sin 90°. Because sin 90° = 1, n1 sin ic = n2, so sin ic = n2 ÷ n1.
If we assume that the light ray is passing from a medium into air, then n2 = 1.
(d)
normal
air
n2
water
n1
i > ic
r =i
FIGURE 10.2.4 Light refracts away from the
normal as it moves from water into air as shown
in diagrams (a) and (b). In diagram (c) the angle
of incidence is the critical angle, and the light
is refracted along the interface between the
water and the air. In diagram (d) the light is
undergoing total internal reflection.
CHAPTER 10 | RAY MODEL OF LIGHT
281
For a light ray passing from medium x (with refractive index nx) into air or a
vacuum, the critical angle ic is given by:
sin ic = n1x
Worked example 10.2.4
CALCULATING CRITICAL ANGLE
Calculate the critical angle for light passing from water (n = 1.33) into air.
FIGURE 10.2.5 Optical fibres transmit light using
total internal reflection.
Thinking
Working
Recall the equation for the critical
angle.
sin ic =
Substitute the refractive indexes of
water and air into the formula. (Unless
otherwise stated, assume that the
second medium is air with n2 = 1.)
1
sin ic = 1.33
= 0.7519
Solve for ic.
ic = sin−1 0.7519
1
nx
= 48.8°
PA
3.4
WS
3.8
PA
3.4
Worked example: Try yourself 10.2.4
CALCULATING CRITICAL ANGLE
Calculate the critical angle for light passing from diamond (n = 2.42) into air.
DISPERSION
FIGURE 10.2.6 Dispersion occurs because
each colour of light has a slightly different
refractive index.
When white light passes from one material to another, each colour travels at a
slightly different speed in the new medium and therefore each colour is refracted by
a slightly different amount. In effect, each colour of light has a different refractive
index in a material. This means that when a beam of white light travels from air
into another medium like glass or water, the white light splits up into a spectrum of
colours (Figure 10.2.6).
This phenomenon is known as dispersion. It is responsible for the rainbows
that you see on rainy days or when spraying water with a hose.
PHYSICSFILE ICT
Where does colour come from?
In the 17th century many people believed that white light was ‘stained’ by its interaction
with earthly materials. Isaac Newton very neatly disproved this with a simple experiment
using two prisms (Figure 10.2.7)—one to split light into its component colours and the
other to turn it back into white light. This showed that the various colours were intrinsic
components of white light since, if
colour was a result of ‘staining’, the
second prism should have added more
colour rather than less.
Newton was the first to identify the
colours of the spectrum, which he
described as red, orange, yellow,
green, blue, indigo and violet. He
FIGURE 10.2.7 Newton’s double prism
invented the colour ‘indigo’ because,
experiment showed that white light is made up of
for religious reasons, he thought that
its component colours.
there should be seven colours.
282
MODULE 3 | WAVES AND THERMODYNAMICS
10.2 Review
SUMMARY
• The amount of refraction of a ray of light can be
calculated using Snell’s law:
• Refraction is the change in the direction of light
that occurs when light moves from one medium
to another.
n1 sin i = n2 sin r , where i is the angle of incidence
and r is the angle of refraction.
• Refraction is caused by changes in the speed of
light rays.
• Total internal reflection occurs when the angle of
refraction exceeds 90°.
• The refractive index, nx, of material x is given by
the formula:
nx = c/vx where c is the speed of light in a vacuum
and vx is the speed of light in material x.
• The critical angle, ic, of a material in air can be
calculated using:
sin ic = n1
x
• When light moves from one material to another,
the changes in speed can be calculated using:
n1v1 = n2v2.
KEY QUESTIONS
1
Choose the correct response from those in bold to
complete the sentences about the refractive indexes of
different types of water.
Although pure water has a refractive index of 1.33,
the salt content of seawater makes its refractive index
a little higher at 1.38. Therefore, the speed of light in
seawater is faster than/slower than/the same as in
pure water.
2
Calculate the speed of light in seawater that has a
refractive index of 1.38.
3
Light travels at 2.25 × 108 m s−1 in water and
2.29 × 108 m s−1 in ice. Water has a refractive
index of 1.33. Use this information to calculate the
refractive index of ice.
4
Light travels from water (n = 1.33) into glass (n = 1.60).
The incident angle is 44°. Calculate the angle of
refraction.
5
For the following situations, can total internal reflection
occur?
Incident medium
Refracting medium
a air (n = 1.00)
glass (n = 1.55)
b glass (n = 1.55)
air (n = 1.00)
water (n = 1.33)
c glass (n = 1.55)
d glass (n = 1.55)
glass (n = 1.58)
CHAPTER 10 | RAY MODEL OF LIGHT
283
10.3 Curved mirrors and lenses
It is a relatively simple matter to predict what will happen when a light ray strikes a
flat surface by using either the law of reflection or Snell’s law. When the reflecting
or refracting surfaces are curved, the situations become much more complex. For
centuries, scientists have been using the special properties of curved mirrors and
lenses to create optical instruments that have allowed us to explore the farthest
reaches of the universe.
CURVED MIRRORS
Curved mirrors can be classified as either converging or diverging. A converging
mirror causes parallel rays of light striking it to be focused to a point in front of the
mirror known as the focus or focal point. A diverging mirror causes parallel rays of
light to spread out away from a virtual focus behind the mirror. Concave mirrors
are converging and convex mirrors are diverging. Figure 10.3.1 shows that a curved
mirror can be considered to be a series of tiny plane mirrors arranged in a curve; the
law of reflection (i.e. i = r) applies at each point on the mirror’s surface.
(a)
(b)
FIGURE 10.3.1 Each ray obeys the law of reflection, resulting in (a) converging rays or (b) diverging
rays. A concave mirror has a real focus. The focus for the convex mirror is virtual, since its position is
determined by extrapolating the reflected rays behind the mirror.
The distance from the mirror to the focus is known as the focal length, f, of the
mirror. It depends on the radius of curvature of the mirror. For a spherical mirror,
the focal point of the mirror is half-way between the centre of curvature of the
mirror and its surface (Figure 10.3.2).
Concave mirror
is a portion of
a sphere.
principal axis
r, radius of curvature
P, pole
C, centre of curvature
F, principal focus
f, focal length
FIGURE 10.3.2
284
MODULE 3 | WAVES AND THERMODYNAMICS
Terminology used to describe spherical mirrors.
The focal length of a mirror is usually one-half of its radius of curvature:
f = 12 r
where:
r is the radius of curvature of the mirror (in m)
f is the focal length of the mirror (in m).
PHYSICS IN ACTION
WE
Finding the focal length of a
concave mirror
The focal length of a concave mirror
can be found experimentally by
focusing the image of a distant
object onto a screen consisting of a
piece of white card or paper. When
the image is clearly focused, the
distance between the mirror and
the screen is the focal length of
the mirror.
concave
mirror
image
focal
length
distant
object
screen
FIGURE 10.3.3 Finding the focal length of
a concave mirror.
Ray diagrams
Ray diagrams are often used to predict the sort of image that will be produced by a
curved mirror in a particular situation. Figure 10.3.4 shows some examples of ray
diagrams for concave mirrors.
Ray diagrams must be drawn to scale if they are to be useful. The horizontal
scale used on the diagram may be different to the vertical scale. The position of the
image is found by drawing the following rays starting at the top of the object being
viewed in the mirror:
• Ray 1: A ray parallel to the principal axis that is reflected from the mirror to
pass through the focus (Figure 10.3.4a)
• Ray 2: A ray passing through the centre of curvature of the mirror that is
reflected straight back along itself (Figure 10.3.4b)
• Ray 3: A ray striking the pole that is reflected as from a plane mirror (Figure 10.3.4c)
• Ray 4: A ray passing through the focus that is reflected to travel back parallel to
the principal axis (Figure 10.3.4d).
(a) Ray 1
(c) Ray 3
principal
axis
object
F
(b) Ray 2
object
i
r
P
(d) Ray 4
object
object
C
F
The path of four rays leaving the top of an object reflect from the mirror to produce an
image. (a) This ray travels parallel to the principal axis and reflects through F. (b) Ray 2 passes through
C and is reflected back along its incident path. (c) The ray striking the pole will reflect as if it has struck
a small plane mirror, i = r. (d) The ray passing through F reflects parallel to the principal axis.
FIGURE 10.3.4
CHAPTER 10 | RAY MODEL OF LIGHT
285
If accurately constructed, these rays should cross at a single point or form a
small triangle. This is shown in Figure 10.3.5. Note that in this diagram only two
of the four rays described above have been drawn. While this is all that is needed to
locate the image, it is good practice to also draw the other two rays to check that they
pass through the same point.
Once the top of the image has been located, the image is then represented as
a vertical arrow starting at the principal axis. The position and orientation of the
image will depend on the type of mirror (i.e. convex or concave), the focal length of
the mirror, and the distance between the object and the mirror.
object
image
In a ray diagram, the point of
intersection of the reflected rays locates the top
of the image.
FIGURE 10.3.5
Concave mirrors
Figure 10.3.6 shows that the type of image formed by a concave mirror depends
on the distance between the object and the mirror relative to the focal length of
the mirror.
(a)
(b)
object
C
Image
real
diminished
inverted
F
image
object
C
F
Image
real
same size
inverted
image
object
(c)
Image
real
magnified
inverted
F
C
image
object
(d)
C
object
(e)
Image
formed
at infinity
F
C
F
Image
virtual
magnified
upright
A concave mirror will produce (a)–(c) a real image when the object is placed farther than the focal point, and (e) a virtual
image when the object is at, or closer than, the focal length. In (d) no image is created becauses the light rays are parallel and do
not converge.
FIGURE 10.3.6
An image formed on the same side of the mirror as the object are known as a
real image. It can be observed by placing a screen at the position indicated. Real
images formed by concave mirrors can be either enlarged or diminished depending
on the position of the object. Real images formed by a concave mirror are always
inverted (i.e. they appear upside-down relative to the object). When the object
is placed closer to the mirror than the focal length, an enlarged, virtual image is
formed. This appears behind the mirror and can only be observed by looking into
the mirror.
286
MODULE 3 | WAVES AND THERMODYNAMICS
Convex mirrors
Regardless of where the object is placed relative to a convex mirror, it will always
produce an upright, diminished, virtual image (Figure 10.3.7). The position of the
object simply determines the size and position of the image.
(a)
object
image F
C
image F
C
F
C
(b)
object
(c)
object
image
Ray tracing for a convex mirror. All images are upright, virtual and diminished. As the
object is brought closer to the mirror, the image increases in size, but it will never be the same size as
the object.
FIGURE 10.3.7
THE MIRROR FORMULA
The position of the image of an object viewed using a convex or concave mirror can
also be determined using the mirror formula.
1
f
= 1u + 1v
where:
f is the focal length of the mirror
u is the distance between the object and the mirror
v is the distance between the image and the mirror.
When using the mirror formula, all distances are measured in the same unit
(usually cm).
TABLE 10.3.1 Summary of negative and positive
results for the focal length and image distance in
curved mirrors.
Negative
This formula assumes that the focal length of a concave mirror will be given as a
positive value and the focal length of a convex mirror is negative. A negative answer
for the image distance means that the image is virtual. A positive answer implies a
real image (see Table 10.3.1).
Positive
Focal length
Convex
mirror
Concave
mirror
Image
distance
Virtual
image
Real image
CHAPTER 10 | RAY MODEL OF LIGHT
287
Worked example 10.3.1
LOCATING THE IMAGE FORMED BY A CONCAVE MIRROR
An object is placed 1.5 m in front of a concave mirror with a focal length of 50 cm.
a Construct a ray diagram of this situation to describe the type of image formed.
Thinking
Working
Construct the ray diagram. (Only two
rays are needed.)
1.5 m
object
image
F
0.5 m
optical
axis
Interpret the ray diagram.
The image is real, inverted and
diminished.
b Use the mirror formula to locate the image.
Thinking
Working
Recall the mirror formula.
1
f
Substitute the information from the
question into the formula.
1
50
Solve the equation for v.
1
v
1
u
=
+ 1v
1 +1
= 150
v
=
1
50
1 = 3 − 1 = 2
− 150
150 150
150
∴ v = 150
= 75 cm
2
Interpret the answer. A negative
answer for the image distance means
that the image is virtual. A positive
answer implies a real image.
Since the answer is positive, this
confirms that the image is a real
image. It will be formed 75 cm away
from the mirror.
Worked example: Try yourself 10.3.1
LOCATING THE IMAGE FORMED BY A CONCAVE MIRROR
A dentist wishing to view a cavity in a tooth holds a concave mirror of focal length
12 mm at a distance of 8 mm from the tooth.
optical
axis
F
(12 mm)
object
(8 mm)
a Construct a ray diagram of this situation to describe the type of image formed.
b Use the mirror formula to locate the image.
288
MODULE 3 | WAVES AND THERMODYNAMICS
Worked example 10.3.2
LOCATING THE IMAGE FORMED BY A CONVEX MIRROR
A shop uses a convex mirror of focal length 2.0 m for security purposes.
If a person 1.5 m tall is standing 4.0 m from the mirror, describe the nature of the
image seen.
a Construct a ray diagram of this situation to describe the type of image formed.
Thinking
Working
Construct the ray diagram. (Only two
rays are needed.)
ray 1
image
ray 2
F
object
Interpret the ray diagram.
C
The image is virtual, upright and
diminished.
b Use the mirror formula to locate the image.
Thinking
Working
Recall the mirror formula.
1
f
Substitute the information from the
question into the formula.
Note that the focal length is negative for
a convex mirror.
1 = 1 +1
v
−2.0
4.0
Solve the equation.
1
v
=
1
u
= − 12 − 14 =
∴v =
Interpret the answer.
+ 1v
−4
3
−2
4
− 14 =
−3
4
= −1.3 m
Since the answer is negative, the image
is a virtual image that appears to be
located 1.3 m behind the mirror.
CHAPTER 10 | RAY MODEL OF LIGHT
289
Worked example: Try yourself 10.3.2
LOCATING THE IMAGE FORMED BY A CONVEX MIRROR
An object is placed 30 cm in front of a convex mirror of focal length 15 cm.
Determine the position of the image.
a Construct a ray diagram of this situation to describe the type of image formed.
b Use the mirror formula to locate the image.
Although using the formula is usually much quicker and more accurate than
using a ray diagram, it is good practice to make a quick sketch of the ray diagram
for any particular problem before you apply the formula. This will give you an idea
of what sort of answer to expect and help you to identify errors in your calculations.
MAGNIFICATION
The size of an image is often expressed in terms of magnification. Magnification
is the ratio between the size of the image and the size of the object. For example, a
magnification of 2 means that the image is twice the size of the object; a magnification
of 0.1 means that the image is one-tenth the size of the object. The magnification
of an image can also be shown to be equal to the ratio of the image distance to the
object distance.
h
M = hi = uv
o
where:
M is the magnification
hi is the height of the image
ho is the height of the object
u is the distance between the object and the mirror
v is the distance between the image and the mirror.
As with the mirror formula, all distances are measured in the same unit (usually cm).
Since magnification is a ratio, it is dimensionless, i.e. it has no units.
Worked example 10.3.3
USING THE MAGNIFICATION FORMULA
A concave mirror forms a real image of an object placed 10 cm in front of the
mirror. The image forms at a distance of 20 cm from the mirror.
If the object is 5 cm high, find the magnification and height of the image.
290
Thinking
Working
Recall the magnification formula.
M=
hi
ho
=
v
u
Substitute the information from the
question into the formula.
M=
hi
5
=
20
10
Solve the equation.
M = 2 and hi = 10 cm
Interpret the answer.
The image is magnified 2 times, so it
appears to be 10 cm high.
MODULE 3 | WAVES AND THERMODYNAMICS
Worked example: Try yourself 10.3.3
USING THE MAGNIFICATION FORMULA
An object 12 cm high is placed 30 cm in front of a convex mirror of focal length
15 cm. The mirror produces a virtual image 10 cm behind the mirror.
converging lenses
Calculate the magnification and height of the image.
LENSES
A lens is a piece of transparent material that has been shaped in order to cause the
light passing through it to bend in a particular way. Lenses can take on a variety
of shapes depending on their function and the exact optical properties required of
them (Figure 10.3.8). Most common lenses have spherical curvature. This means
that each face is part of a sphere. Other shapes are possible but are much more
difficult to achieve. These are usually used only in expensive optical instruments.
We will refer only to bi-concave or bi-convex lenses. (Since these are by far the
most common types of lens, scientists often just call these ‘concave’ and ‘convex’
lenses respectively.)
Like curved mirrors, lenses can be classified as either converging or diverging
depending on their effect on parallel rays of light. Convex lenses are converging and
concave lenses are diverging (Figure 10.3.9).
(a)
FIGURE 10.3.9
plano-convex
bi-convex
diverging lenses
plano-concave
bi-concave
FIGURE 10.3.8 Convex lenses are thicker in
the middle. Concave lenses are thinner in the
middle. Plano-convex and plano-concave lenses
each have one flat surface.
(b)
(a) A convex lens causes light to converge. (b) A concave lens causes light to diverge.
In the case of a concave lens, the curvature of the lens faces means that parallel
rays diverge after passing through the lens(a)
(Figure 10.3.10b). The diverging rays
can be traced back through the lens to a point behind the lens. This is the principal
focus for a concave lens. It is sometimes known as a virtual focus since the rays only
F
meet at this point in our imagination. A virtual image formed by a lens can only be
seen when looking directly through the lens.
(b)
(a)
F
F
FIGURE 10.3.10 A beam of light can be considered as a bundle of separate rays. Each ray refracts
(b)
as it enters and leaves the lens. This causes the rays to (a) converge or (b) diverge depending on the
shape of the lens.
F
CHAPTER 10 | RAY MODEL OF LIGHT
291
(b) principal Each
focus lens is considered to have two focal points (foci), one either side of the lens.
(a)
principal focus
principal
axis
f
optical centre
Ray diagrams
(b) principal focus
cus
f
principal
axis
centre
optical centre
Terminology associated with
spherical lenses. (a) If rays parallel to the
principal axis enter a convex lens, they converge
to the principal focal point, f. (b) For a concave
lens, rays parallel to the principal axis diverge as
if they come from a virtual focus.
FIGURE 10.3.11
Each lens also has a special point known as its optical centre. A light ray that passes
through the optical centre of the lens experiences no refraction (i.e. it passes straight
principalThe distance between the optical centre and the focal
through the lens undeviated).
axis
f
point is known
as the focal
length of the lens. As with mirrors, it is helpful to imagine
the principal axis of the lens, which is a line that passes through the two focal points
and theoptical
optical
centre (Figure 10.3.11).
centre
Just as with mirrors, it is helpful to create a ray diagram to better understand how
images are formed through lenses. Three separate rays are generally sufficient to
determine the location of the image:
• Ray 1: A ray passing through the optical centre of a lens is not deflected
(Figure 10.3.12a).
• Ray 2: A ray parallel to the principal axis is refracted such that it passes through
the principal focus (Figure 10.3.12b).
• Ray 3: A ray passing through the focus is refracted to travel parallel to the
principal axis (Figure 10.3.12c).
(a)
(b)
F
F
(c)
F
F
(a) A ray passing through the optical centre is undeviated. (b) A ray parallel to the
principal axis passes (or appears to pass) through the principal focal point after passing through the
lens. (c) A ray which passes through the focal point emerges parallel to the principal axis.
FIGURE 10.3.12
Figures 10.3.13 and 10.3.14 demonstrate how ray diagrams can be used to
locate the images of objects at various distances from convex and concave lenses.
292
MODULE 3 | WAVES AND THERMODYNAMICS
(a)
object
Description
of image
F
2F
(b)
F
real
inverted
diminished
2F
image
object
F
2F
real
inverted
same size
2F
F
image
(c)
(a)
F image
object 2F
(b)
object
2F
F image
2F
object F image
object
F
2F
real
inverted
enlarged
F
2F
(c)
image
(d)
object
F
2F
2F
F
image
at infinity
(d)
object
(e)
image
2F
object
2F
F
F
2F
virtual
upright
enlarged
(a)–(c) When the object is placed beyond the focal point of a convex lens, a real
image is formed. As the object is brought closer, a larger image is created. (d) The image of an object
at the focal point of a convex lens is considered to be at infinity. (e) A convex lens forms an upright,
virtual image of an object located inside the focal point.
(e)
object
FIGURE 10.3.13
The lens formula
The position and nature of the image formed by a lens can also be calculated using
the lens formula:
1
f
F image
2F
F
image
A concave lens causes rays to
diverge and a virtual, reduced image is always
formed.
FIGURE 10.3.14
= 1u + 1v
This formula is exactly the same as the mirror formula and the sign conventions
are similar: the focal length of a diverging concave lens is given a negative value and
the focal length of a convex lens is positive. A negative value for the image distance
means that the image is virtual, and a positive value implies a real image (Table 10.3.2).
Magnification
Just as for curved mirrors, the size of an image in a lens can be expressed in terms
of magnification. Magnification is the ratio between the size of the image and the
size of the object. The magnification formula for lenses is the same as for mirrors.
h
M = hi = uv
o
TABLE 10.3.2 Summary of negative and positive
results for the focal length and image distance in
curved lenses.
Negative
Positive
Focal length
Concave lens
Convex lens
Image
distance
Virtual
image
Real image
CHAPTER 10 | RAY MODEL OF LIGHT
293
Worked example 10.3.4
USING THE LENS AND MAGNIFICATION FORMULAS
A child decides to examine a beetle by using a convex lens of focal length 7 cm by
holding the magnifying glass 14 cm from the beetle.
a Construct a ray diagram to describe the image.
Thinking
Working
Construct the ray diagram.
(Only two rays are needed.)
ray 2
ray 1
F
f = 7 cm
F
u = 14 cm
Interpret the ray diagram.
The image is real, inverted and the same size as
the object.
b Use the lens formula to locate the image.
Thinking
Working
Recall the lens formula.
1
f
Substitute the information
from the question into
the formula.
Note that the focal length is positive for a
convex lens.
1
7
1 +1
= 14
v
Solve the equation.
1
v
=
=
1
u
+ 1v
1
7
1 = 2 − 1 = 1
− 14
14 14
14
v = 14 cm
Interpret the answer.
Because the answer is positive, the image is a
real image 14 cm from the lens.
c If the beetle is 2.5 cm long, calculate the size of the image.
294
Thinking
Working
Recall the magnification
formula.
M=
hi
ho
=
Substitute the information
from the question into the
formula.
M=
hi
2.5
= 14
14
Solve the equation.
M = 1 and hi = 2.5 cm
Interpret the answer.
The image is the same size as the object and
appears 2.5 cm long.
MODULE 3 | WAVES AND THERMODYNAMICS
v
u
Worked example: Try yourself 10.3.4
USING THE LENS AND MAGNIFICATION FORMULAS
An artist uses a concave lens of focal length 10 cm to reduce a sketch of a tree to
see what it would look like in miniature. The tree in the sketch is 20 cm tall. The
lens is held 15 cm above the page.
a Construct a ray diagram to describe the image.
b Use the lens formula to determine the image distance and whether the image
formed is real or virtual.
WS
3.6
c Calculate the magnification of the image.
PA
3.3
PA
3.3
10.3 Review
SUMMARY
• The focal point of a mirror or lens is the point at
which parallel rays are focused.
• Ray diagrams can be used to determine the nature
of the image formed by a curved mirror or lens.
• Concave mirrors are converging mirrors; convex
mirrors are diverging mirrors.
• Mirror/lens formula:
• Convex lenses are converging lenses; concave
lenses are diverging lenses.
• Magnification formula: M =
1
f
=
1
u
+ 1v
hi
ho
=
v
u
KEY QUESTIONS
1
aWhat is meant by the term ‘radius of curvature’ for
a spherical mirror?
6
A convex lens of focal length 20 cm is used to form
an image of the full moon on a screen. The moon is
at a distance of 3.9 × 108 m from the Earth and has a
diameter of 3.5 × 106 m. How far must the screen be
placed from the lens?
7
An object is 10 cm from a convex lens of focal length
15 cm. The image formed is:
A real, inverted and enlarged
B real, inverted and the same size as the object
C virtual, upright and reduced
D virtual, upright and enlarged.
bIf the radius of curvature of a spherical concave
mirror is 50 cm, what is its focal length?
2
What happens to light rays as they are reflected from a
concave mirror compared to a convex mirror?
3
A pen is held 60 cm from a convex mirror of focal
length 20 cm. Is the image:
a real or virtual?
b upright or inverted?
c enlarged, diminished or the same size?
4
A concave mirror of focal length 20 cm is used to
produce an image of an object. How far from the
mirror must the object be placed in order to form an
image that is inverted and the same size as the object?
5
A cavity in a tooth is 2 mm long. A dentist uses a
concave mirror of focal length 15 mm to view the
tooth. The mirror is held 8 mm from the tooth.
a Describe the nature of the image formed.
b Approximately how big will the cavity appear?
CHAPTER 10 | RAY MODEL OF LIGHT
295
Chapter review
KEY TERMS
centre of curvature
critical angle
diffuse
dispersion
focal length
focus
intensity
magnification
radius of curvature
ray diagram
real image
refraction
refractive index
Snell’s law
total internal reflection
virtual image
REVIEW QUESTIONS
1
Recall the law of reflection.
2
Explain the difference between a real image and a
virtual image.
3
A person stands 0.75 m in front of a plane mirror.
a Describe the nature of the image of themselves that
they will see.
b Calculate the distance between the person and their
image.
4
A 1.6 m tall person stands 1 m in front of a vertical
dressing mirror. The person’s eyes are at a height of
1.5 m above the ground.
a How high above the ground should the bottom of
the mirror be positioned so that the person can still
see their feet? (Hint: Draw a ray diagram showing
the ray from the person’s feet to their eyes.)
b How high above the ground should the mirror
extend so that the person can see the top of their
head?
c What is the minimum length of the mirror that will
allow the person to see their entire body?
5
If the distance between a light source and an observer
is halved, how much will the intensity of the light be
changed?
6
A projector is creating an image on a screen at the
front of a classroom. For a student sitting 3.0 m away
from the screen, the image has an intensity of 5.0 cd.
What would be the intensity for a student sitting 4.5 m
away from the screen?
7
Choose the correct answers from those given in bold
to complete the following sentence about refraction.
As light travels from quartz (n = 1.46) to water
(n = 1.33), its speed increases/decreases which
causes it to refract away from/towards the normal.
8
296
The figure below represents a situation involving the
refraction of light. Identify the correct label for each
of the lines from the choices provided: boundary
between media, reflected ray, incident ray, normal,
refracted ray.
MODULE 3 | WAVES AND THERMODYNAMICS
b
a
c
medium 1
medium 2
d
e
9
Light passes from glycerine (nglycerine = 1.47) into water
(nwater = 1.33) with an angle of incidence of 23.0°.
Calculate the angle of refraction.
10 The speed of light in air is 3.00 × 108 m s−1. As light
strikes an air–perspex boundary, the angle of incidence
is 43.0° and the angle of refraction is 28.5°. Calculate
the speed of light in perspex.
11 A ray of light travels from air, through a layer of glass
and then into water as shown in the figure. Calculate
angles a, b and c.
air
(n = 1.00)
glass
(n = 1.50)
water
(n = 1.33)
40°
a
b
c
12 A ray of light exits a glass block. On striking the inside
wall of the glass block the ray makes an angle of 58.0°
with the glass–air boundary. The index of refraction of
the glass is 1.52. Calculate the:
a angle of incidence
b angle of refraction of the transmitted ray (assuming
nair = 1.00)
c angle of deviation (i.e. the difference between the
angles of incidence and refraction)
d speed of light in the glass.
13 A narrow beam of white light enters a crown glass
prism with an angle of incidence of 30.0°. In the
prism, the different colours of light are slowed to
varying degrees. The refractive index for red light in
crown glass is 1.50 and for violet light the refractive
index is 1.53. Calculate the:
a angle of refraction for the red light
b angle of refraction for the violet light
c angle through which the spectrum is dispersed
d speed of the violet light in the crown glass.
14 Calculate the critical angle for light travelling between
the following media.
Incident medium
Refracting medium
a
ice (n = 1.31)
air (n = 1.00)
b
salt (n = 1.54)
air (n = 1.00)
c
cubic zirconia (n = 2.16)
air (n = 1.00)
15 When a light ray refracts, the difference between
the angle of incidence and angle of refraction is
known as the angle of deviation. Sort the following
boundaries between media in order of increasing
angle of deviation.
A water (n = 1.33) to diamond (n = 2.42)
B water (n = 1.33) to air (n = 1.00)
C air (n = 1.00) to diamond (n = 2.42)
D glass (n = 1.50) to air (n = 1.00)
16 In 1905, Albert Einstein proposed that nothing in the
universe can travel faster than the speed of light in
a vacuum. Assuming that this is correct, what is the
minimum refractive index that any medium can have?
17 A person wants to use a concave mirror with a focal
length of 40 cm as a make-up mirror. She stands too
far back and sees an image of her face that is true to
size but inverted.
a How far is her face from the mirror?
b Where should she stand to see an upright,
magnified image of herself?
18 A person looks at a convex mirror of focal length 3 m
which is placed at a blind corner. He sees a truck
coming.
a Describe the nature of the image when the truck is
30 m from the mirror.
b What is the magnification of the truck?
19 Locate the images formed when a match 3 cm long
is placed at the following distances from a concave
mirror of focal length 5 cm.
a 10 cm
b 5 cm
c 2 cm
20 Explain why the image formed by a concave lens is
always a virtual image.
21 A real image is formed 1.2 m from a convex lens of
focal length 0.4 m. The object is 5 cm tall. Determine:
a the distance of the object from the lens
b the height of the image.
22 A child examines an insect with a magnifying glass of
focal length 15 cm. The insect is 5 mm in length and
10 cm from the lens.
a How far is the image from the lens?
b What magnification occurs?
c How large is the image of the insect?
23 A concave lens is held 15 cm above a page and the
text appears to be one-third its original size.
a Is the image real or virtual?
b Approximately how far from the lens is the image
formed?
c What is the focal length of the concave lens?
24 A person wants to use a lens to project a magnified
image from a small slide onto a screen.
a Should the lens be concave or convex?
b What should the focal length of the lens be if the
slide will be placed 4.0 cm away from the lens and
the image needs to be 30 times larger than the
original slide?
25 After completing the activity on page 278, reflect
on the inquiry question: What properties can be
demonstrated when using the ray model of light? In
your response, discuss the concept of refraction.
CHAPTER 10 | RAY MODEL OF LIGHT
297
CHAPTER
Thermodynamics
Due to increasing levels of carbon dioxide in the atmosphere, the Earth is getting
warmer. The last two decades of the twentieth century were the warmest for more than
400 years. In 2014 the Earth was the warmest since records began in 1890. These
higher temperatures increase the severity of bushfires. The rate of evaporation from
pastures also increases, drying the land and reducing the level of food production.
Thermal energy is part of our everyday experience. Humans can thrive in the
climatic extremes of the Earth, from the outback deserts to ski slopes in winter.
Content
INQUIRY QUESTION
How are temperature, thermal energy and particle motion related?
By the end of this chapter you will be able to:
• explain the relationship between the temperature of an object and the kinetic
energy of the particles within it (ACSPH018)
• explain the concept of thermal equilibrium (ACSPH022)
• analyse the relationship between the change in temperature of an object and its
specific heat capacity through the equation ∆Q = mc ∆T (ACSPH020)
• investigate energy transfer by the process of:
-- conduction
-- convection
-- radiation (ACSPH016)
• conduct an investigation to analyse qualitatively and quantitatively the latent heat
involved in a change of state
• model and predict quantitatively energy transfer from hot objects by the process
of thermal conductivity CCT
• apply the following relationships to solve problems and make quantitative
predictions in a variety of situations: ICT N
-- ∆Q = mc ∆T , where c is the specific heat capacity of a substance
--
Q
t
=
kA∆T
d
, where k is the thermal conductivity of a material.
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
11.1 Heat and temperature
PHYSICS INQUIRY
CCT ICT
Temperature and thermal energy
How are temperature, thermal energy and particle motion
related?
COLLECT THIS…
•
9 large marbles
•
9 small marbles
•
2 lots of 9 small compression springs, each lot a different stiffness
•
sheet of cardboard
•
hot glue gun
DO THIS…
1 Using the hot glue gun, glue the springs to the marbles in a line. Create two
different lines, one with large marbles and one with small marbles.
2 Fold the cardboard to create a channel wide enough to hold both lines of
marbles side by side with a gap in between. Cut out a strip of cardboard
and glue it to the bottom of the channel to create 2 side-by-side channels
for the different lines of marbles.
3 Glue the first marble of each line to the bottom of the cardboard channel.
4 Starting with all the marbles stationary, add a quick pulse-like movement to
the cardboard. Observe how the marbles move.
5 Try shaking the cardboard channel with a constant frequency and small
amplitude. Once all the marbles are vibrating try suddenly changing the
frequency to be faster.
6 Repeat step 5 but this time with a slower frequency.
RECORD THIS…
The marbles represent the molecules in a solid. When vibrating, these marbles
have a kinetic energy. The faster they are moving, the more kinetic energy they
have. In this model, the kinetic energy represents temperature.
You are adding energy to the system when you shake the cardboard. This
represents thermal energy. The faster the cardboard shakes the more thermal
energy that represents.
Describe the different things that can affect particle motion in your model.
Present your results by creating a diagram showing how your marble set-up
links with thermal energy, kinetic energy and temperature.
REFLECT ON THIS…
How are temperature, thermal energy and particle motion related?
This model shows how heat is conducted in a solid. How could you show how
thermal energy (heat) is transferred from a hot object to a cold object made
from different materials?
In the 16th century, Francis Bacon, an English essayist and philosopher, proposed a
radical idea: that heat is motion. He went on to write that heat is the rapid vibration of
tiny particles within every substance. At the time, his ideas were dismissed because
the nature of particles wasn’t fully understood. An opposing theory at the time was
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that heat was related to the movement of a fluid called ‘caloric’ that filled the spaces
within a substance.
Today, it is understood that all matter is made up of small particles (atoms or
molecules). Using this knowledge, it is possible to look more closely at what happens
during heating processes.
This section starts by looking at the kinetic particle model, which states that
the small particles (atoms or molecules) that make up all matter have kinetic energy.
This means that all particles are in constant motion, even in extremely cold solids.
It was thought centuries ago that if a material was continually made cooler, there
would be a point at which the particles would eventually stop moving. This coldest
possible temperature is called absolute zero and is discussed later in this section.
KINETIC PARTICLE MODEL
Much of our understanding of the behaviour of matter today depends on a model
called the kinetic particle model (kinetic theory).
Recall that a model is a representation that describes or explains the workings
within an object, system or idea. These are the assumptions behind the kinetic
particle model:
•All matter is made up of many very small particles (atoms or molecules).
• The particles are in constant motion.
• No kinetic energy is lost or gained overall during collisions between particles.
• There are forces of attraction and repulsion between the particles in a
material.
• The distances between particles in a gas are large compared with the size of
the particles.
The kinetic theory applies to all states (or phases) of matter: solids, liquids
and gases.
(a)
Solids
Within a solid, the particles must be exerting attractive forces or bonds on each
other for the matter to hold together in its fixed shape. There must also be repulsive
forces. Without the repulsive forces, the attractive forces would cause the solid
to collapse. In a solid, the attractive and repulsive forces hold these particles in
more or less fixed positions, usually in a regular arrangement or lattice shown in
Figure 11.1.1a. In a solid, the particles are not completely still; they vibrate around
average positions. The forces on individual particles are sometimes predominantly
attractive and sometimes repulsive, depending on their exact position relative to
neighbouring particles.
(b)
Liquids
Within a liquid, there is still a balance of attractive and repulsive forces. Compared
with a solid, the particles in a liquid have more freedom to move around each other
and therefore take the shape of the container (Figure 11.1.1b). Generally, the liquid
takes up a slightly greater volume than it would in the solid state. Particles collide
but remain attracted to each other, so the liquid remains within a fixed volume but
with no fixed shape.
(c)
Gases
In a gas, particles are in constant, random motion, colliding with each other and the
walls of the container. The particles move rapidly in every direction, quickly filling
the volume of any container and occasionally colliding with each other as shown
in Figure 11.1.1c. A gas has no fixed volume. The particle speeds are high enough
that, when the particles collide, the attractive forces are not strong enough to keep
the particles close together. The repulsive forces cause the particles to separate and
move off in other directions.
FIGURE 11.1.1 (a) Molecules in a solid have
low kinetic energy and vibrate around average
positions within a regular arrangement. (b) The
particles in a liquid have more kinetic energy
than those in a solid. They move more freely
and take the shape of the container. (c) Gas
molecules are free to move in any direction.
CHAPTER 11 | THERMODYNAMICS
301
PHYSICSFILE ICT
States of matter: plasma
The three basic phases (states) of matter are solid, liquid and gas. These are
generally all that are discussed in secondary science.
There are in fact four phases of matter that are observable in everyday life—solid,
liquid, gas and plasma. Under special conditions, several more exist. Plasma exists
when matter is heated to very high temperatures and electrons are freed (ionisation).
A gas that is ionised and has an equal number of positive and negative charges is
called plasma. The interior of stars consists of plasma. In fact, most of the matter in
the universe is plasma (Figure 11.1.2).
FIGURE 11.1.2
99.9% of the visible universe is made up of plasma.
THE KINETIC PARTICLE MODEL, INTERNAL ENERGY
AND TEMPERATURE
•Heating is a process that
always transfers thermal
energy from a hotter substance
to a colder substance.
• Heat is measured in joules (J).
• Temperature is related to the
average kinetic energy of the
particles in the substance.
The faster the particles move,
the higher the temperature
of the substance.
302
The kinetic particle model can be used to explain the idea of heat as a transfer of
energy. Heat (measured in joules) is the transfer of thermal energy from a hotter
body to a colder one. Heating is observed by the change in temperature, the change
of state or the expansion of a substance.
When a solid substance is ‘heated’, the particles within the material gain either
kinetic energy (move faster) or potential energy (move away from their
equilibrium positions).
The term heat refers to energy that is being transferred (moved). So it is
incorrect to talk about heat contained in a substance. The term internal energy
refers to the total kinetic and potential energy of the particles within a substance.
Heating (the transfer of thermal energy) changes the internal energy of a substance
by affecting the kinetic energy and/or potential energy of the particles within the
substance. The movement of the particles in a substance due to kinetic energy is
ordered: the particles move back and forth and we can model their behaviour. In
comparison, the internal energy of a system is associated with the chaotic motion of
the particles—it concerns the behaviour of a large number of particles that all have
their own kinetic and potential energy.
MODULE 3 | WAVES AND THERMODYNAMICS
Using the kinetic particle model, an increase in the total internal energy of the
particles in a substance results in an increase in temperature if there is a net gain in
kinetic energy. Hot air balloons are an example of this process in action. The air in
a hot air balloon is heated by a gas burner to a maximum of 120°C. The nitrogen
(78%) and oxygen (21%) molecules in the hot air gain energy and so move a lot
faster. The air in the balloon becomes less dense than the surrounding air, causing
the balloon to float as seen in Figure 11.1.3.
(a)
(b)
FIGURE 11.1.3 (a) Nitrogen and oxygen molecules gain energy when the air is heated, lowering the
density of the air and causing the hot air balloon to rise off the ground. (b) A thermal image shows
the temperature of the air inside the balloon.
Sometimes heating results only in the change of state or expansion of an object, and
not a change in temperature. In these cases, the total internal energy of the particles
has still increased but only the potential energy has increased, not the kinetic energy.
For instance, particles in a solid being heated continue to be mostly held in place,
due to the relatively strong interparticle forces. For the substance to change state
from solid to liquid, it must receive enough energy to separate the particles from
each other and disrupt the regular arrangement of the solid. During this ‘phase
change’ process, the energy is used to overcome the strong interparticle forces, but
not to change the overall speed of the particles. In this situation, the temperature
does not change. This is discussed in more detail in Section 11.3.
MEASURING TEMPERATURE
Only four centuries ago, there were no thermometers and people described heating
effects by vague terms such as hot, cold and lukewarm. In about 1593, Italian
inventor Galileo Galilei made one of the first thermometers. His ‘thermoscope’ was
not particularly accurate as it did not take into account changes in air pressure, but it
did suggest some basic principles for determining a suitable scale of measurement.
His work suggested that there be two fixed points: the hottest day of summer and
the coldest day of winter. A scale like this is referred to as an arbitrary scale, because
the fixed points are randomly chosen.
Celsius and Fahrenheit scales
Two of the better-known arbitrary temperature scales are the Fahrenheit and Celsius
scales. Gabriel Fahrenheit of Germany invented the first mercury thermometer
in 1714. While Fahrenheit is used in the United States of America to measure
temperature, the system used in most countries in the world is the Celsius scale.
Absolute scales are different from arbitrary scales. For a scale to be regarded as
‘absolute’, it should have no negative values. The fixed points must be reproducible
and have zero as the lowest value.
Kelvin scale
When developing the absolute temperature scale, the triple point of water provides
one reliable fixed point. This is a point where the combination of temperature and
air pressure allows all three states of water to coexist. For water, the triple point is
only slightly above the standard freezing point (0.01°C) and provides a unique and
repeatable temperature with which to adjust the Celsius scale.
CHAPTER 11 | THERMODYNAMICS
303
•The freezing point of
water (0°C) is equivalent to
273.15 K (kelvin). This is often
approximated to 273 K.
• The size of each unit (1°C or
1 K) is the same.
• The word ‘degree’ and the
degree symbol are not used
with the kelvin scale.
• To convert a temperature
from degrees Celsius to kelvin,
add 273.
• To convert a temperature
from kelvin to degrees Celsius,
subtract 273.
•0 degrees Celsius (0°C) is
the freezing point of water at
standard atmospheric pressure.
• 100°C is the boiling point of
water at standard atmospheric
pressure.
• Absolute zero = 0 K
= −273.15°C.
• All molecular motion ceases
at absolute zero. This is the
coldest temperature possible.
The absolute or kelvin temperature scale is based on absolute zero and the triple
point of water. See Figure 11.1.4 for a comparison of the kelvin and Celsius scales.
Kelvin
Celsius
boiling point of water
373.15
100.00°
triple point of water
freezing point of water
273.16
273.15
0.01°
0°
absolute zero
0°
̵273.15°
Comparison of the kelvin and Celsius scales. Note that there are no negative values
on the kelvin scale.
FIGURE 11.1.4
Absolute zero
Experiments indicate that there is a limit to how cold things can get. The kinetic
theory suggests that if a given quantity of gas is cooled, its volume decreases. The
volume can be plotted against temperature and results in a straight line graph as
shown in Figure 11.1.5.
THE LAWS OF THERMODYNAMICS
The topic of thermal physics involves the phenomena associated with the energy
transfer between objects at different temperatures. Since the nineteenth century,
scientists have developed four laws for this subject. The first two are studied in this
section.
The first, second and third laws had been known and understood for some time
before another fourth law was determined. This final law was considered to be so
important that it was decided to be placed first, and so it is called the zeroth law of
thermodynamics.
The zeroth law of thermodynamics
Volume, V
ideal gas behviour
freezes to form
a solid
condenses to form
a liquid
–273.15 –200 –100
0
100
Temperature, T (°C)
FIGURE 11.1.5 Gases have smaller volumes
as they cool. This relationship is linear.
Extrapolating (extending) the line to where
the volume is zero gives a theoretical value of
absolute zero.
304
The zeroth law of thermodynamics relates to thermal equilibrium and thermal
contact and allows temperature to be defined.
If two objects are in thermal contact, energy can flow between them. For example,
if an ice cube is placed in a copper pan, the ice molecules are in thermal contact with
the copper atoms. Assuming that the copper is warmer than the ice, thermal energy
flows from the copper to the ice.
Thermal equilibrium is when there is no longer a flow of energy between two
objects in thermal contact. If a frozen piece of steak is placed in a container of warm
water, energy is transferred from the water to the steak. The steak gains energy and
warms up. The water loses energy and cools down. Eventually the transfer of energy
between the steak and water stops. This point is called thermal equilibrium, and the
steak and water are at the same temperature.
The zeroth law combines these concepts. If two objects are in thermal equilibrium
with a third object, then they are all in thermal equilibrium with each other.
MODULE 3 | WAVES AND THERMODYNAMICS
The first law of thermodynamics
The first law of thermodynamics states that energy simply changes from one form
to another and the total internal energy in a system is constant. The internal energy
of the system can be changed by heating or cooling, or by work being done on or by
the system.
Any change in the internal energy (ΔU ) of a system is equal to the energy added
by heating (+Q) or removed by cooling (−Q), minus the work done on (−W ) or by
(+W) the system:
ΔU = Q − W
The internal energy (U ) of a system is defined as the total kinetic and potential
energy of the system. As the average kinetic energy of a system is related to its
temperature and the potential energy of the system is related to the state, then a
change in the internal energy of a system means that either the temperature changes
or the state changes.
If heat (Q) is added to the system, then the internal energy (U ) rises by either
increasing temperature or changing state from solid to liquid or liquid to gas.
Similarly, if work (W ) is done on a system, then the internal energy rises and the
system once again increases in temperature or changes state by melting or boiling.
When heat is added to a system or work is done on a system, ΔU is positive.
If heat (Q) is removed from the system, then the internal energy (U ) decreases
by either decreasing temperature or changing state from liquid to solid or gas
to liquid. Similarly, if work (W ) is done by the system, then the internal energy
decreases and the system once again decreases in temperature or changes state by
condensing or solidifying. When heat is removed from a system or work is done by
a system, ΔU is negative.
If heat is added to the system and work is done by the system, then whether the
internal energy increases or decreases depends on the magnitude of the energy into
the system compared to the magnitude of the energy out of the system.
Worked example 11.1.1
CALCULATING THE CHANGE IN INTERNAL ENERGY
A 1 L beaker of water has 25 kJ of work done on it and also loses 30 kJ of
thermal energy to the surroundings.
If objects A and B are each in
thermal equilibrium with object
C, then objects A and B are in
thermal equilibrium with each
other.
Two objects in thermal
equilibrium with each other must
be at the same temperature.
PHYSICSFILE ICT
Close to absolute zero
As temperatures get close to
absolute zero, atoms start to behave
in weird ways. Since the French
physicist Guillaume Amontons first
proposed the idea of an absolute
lowest temperature in 1699,
physicists have theorised about the
effects of such a temperature and
how it could be achieved. The laws
of physics dictate that absolute zero
itself can be approached but not
reached. In 2003, researchers from
NASA and MIT, in the United States
of America, succeeded in cooling
sodium atoms to one billionth of a
degree above absolute zero. At this
temperature, all elementary particles
merge into a single state, losing their
separate properties and behaving
as a single ‘super atom’, a state
first proposed by Einstein 70 years
earlier.
What is the change in energy of the water?
Thinking
Working
Heat is removed from the system, so Q is negative.
ΔU = Q − W
Work is done on the system, so W is negative.
Note that the units are kJ, so express the final
answer in kJ.
= −30 − (−25)
ΔU = −5 kJ
Worked example: Try yourself 11.1.1
CALCULATING THE CHANGE IN INTERNAL ENERGY
A student places a heating element and a paddle-wheel apparatus in an
insulated container of water. She calculates that the heater transfers 2530 J of
thermal energy to the water and the paddle does 240 J of work on the water.
Calculate the change in internal energy of the water.
CHAPTER 11 | THERMODYNAMICS
305
11.1 Review
SUMMARY
• The kinetic particle theory proposes that all matter
is made of atoms or molecules (particles) that are
in constant motion.
• Heating is a process that always transfers thermal
energy from a hotter substance to a colder
substance.
• In solids, the attractive and repulsive forces hold
the particles in more or less fixed positions,
usually in a regular arrangement or lattice. These
particles are not completely still—they vibrate
about average positions.
• Temperatures can be measured in degrees Celsius
(°C) or kelvin (K).
• In liquids, there is still a balance of attractive and
repulsive forces between particles but the particles
have more freedom to move around. Liquids
maintain a fixed volume.
• To convert from Celsius to kelvin: add 273; to
convert from kelvin to Celsius: subtract 273.
• Absolute zero is called simply ‘zero kelvin’ (0 K)
and it is equal to −273.15°C.
• The size of each unit, 1°C or 1 K, is the same.
• The zeroth law of thermodynamics states that if
objects A and B are each in thermal equilibrium
with object C, then objects A and B are in thermal
equilibrium with each other. A, B and C must be at
the same temperature.
• In gases, the particle speeds are high enough that,
when particles collide, the attractive forces are not
strong enough to keep them close together. The
repulsive forces cause the particles to move off in
other directions.
• The first law of thermodynamics states that energy
simply changes from one form to another and the
total energy in a system is constant.
• Internal energy refers to the total kinetic and
potential energy of the particles within a
substance.
• Any change in the internal energy (ΔU) of a system
is equal to the energy added by heating (+Q) or
removed by cooling (−Q), minus the work done on
(−W) or by (+W) the system: ΔU = Q − W.
• Temperature is related to the average kinetic
energy of the particles in a substance.
KEY QUESTIONS
1
Which of the following is true of a solid?
A Particles are moving around freely.
B Particles are not moving.
C Particles are vibrating in constant motion.
D A solid is not made up of particles.
2
aAn uncooked chicken is placed into an oven
that has been preheated to 180°C. Which of the
following statements describe what happens as
soon as the chicken is placed in the oven? (More
than one answer is possible.)
A Thermal energy flows from the chicken into the
hot air.
B The chicken and the air in the oven are in
thermal equilibrium.
C Thermal energy flows from the hot air into the
chicken.
D The chicken and the air in the oven are not in
thermal equilibrium.
b A chicken is inside an oven that has been preheated
to 180°C. The chicken has been cooking for one
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MODULE 3 | WAVES AND THERMODYNAMICS
hour and its temperature is also 180°C. Which of the
following statements best describes this scenario?
A Thermal energy flows from the chicken into the
hot air.
B The chicken and the air in the oven are in
thermal equilibrium.
C Thermal energy flows from the hot air into the
chicken.
D The chicken and the air in the oven are not in
thermal equilibrium
3
Which of the following temperature(s) cannot possibly
exist? (More than one answer is possible.)
A 1 000 000°C
B −50°C
C −50 K
D −300°C
4
A tank of pure helium is cooled to its freezing point of
−272.2°C. Describe the energy of the helium particles
at this temperature.
5
Convert the following temperatures:
a 30°C into kelvin
b 375 K into degrees Celsius.
6
Sort the following temperatures from coldest to
hottest:
freezing point of water
100 K
absolute zero
−180°C
10 K
7
A hot block of iron does 50 kJ of work on a cold
floor. The block of iron also transfers 20 kJ of
heat energy to the air. Calculate the change in
energy (in kJ) of the iron block.
CHAPTER 11 | THERMODYNAMICS
307
11.2 Specific heat capacity
A small amount of water in a kettle experiences a greater change in temperature
than a larger volume if heated for the same time. A metal object left in the sunshine
gets hotter faster than a wooden object. Large heaters warm rooms faster than
small ones.
These simple observations suggest that the mass, material and the amount of
energy transferred influence any change of temperature.
CHANGING TEMPERATURE
The specific heat capacity of
a material, c, is the amount of
energy that must be transferred to
change the temperature of 1 kg of
the material by 1°C or 1 K.
The temperature of a substance is a measure of the average kinetic energy of the
particles inside the substance. To increase the temperature of the substance, the
kinetic energy of its particles must increase. This happens when heat is transferred
to that substance. The amount the temperature increases depends on a number
of factors.
The greater the mass of a substance, the greater the energy required to change
the kinetic energy of all the particles. So, the heat required to raise the temperature
by a given amount is proportional to the mass of the substance.
ΔQ ∝ m
where:
ΔQ is the heat energy transferred (in J)
m is the mass of material being heated (in kg).
The more heat that is transferred to a substance, the more the temperature of
that substance increases. The amount of energy transferred is therefore proportional
to the change in temperature.
ΔQ ∝ ΔT
where ΔT is the change in temperature in °C or K.
Heating experiments using different materials confirm that these relationships
hold true regardless of the material being heated. However, heating the same
masses of different materials shows that the amount of energy required to heat a
given mass of a material through a particular temperature change also depends on
the nature of the material being heated. For example, a volume of water requires
more energy to change its temperature by a given amount than the same volume
of methylated spirits. For some materials, temperature change occurs more easily
than for others.
Combining these observations, the amount of energy added to or removed from
the substance is proportional to the change in its temperature, its mass and its
specific heat capacity (provided a material does not change state). The specific
heat capacity of a material changes when the material changes state.
Q = mcΔT
where:
Q is the heat energy transferred (in J)
m is the mass (in kg)
ΔT is the change in temperature (in °C or K)
c is the specific heat capacity of the material (in J kg−1 K−1).
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MODULE 3 | WAVES AND THERMODYNAMICS
Table 11.2.1 lists the specific heat capacities for some common materials. You
can see that it also lists the average value for the human body, which takes into
account the various materials within the body and the percentage that each material
contributes to the body’s total mass.
TABLE 11.2.1
Approximate specific heat capacities of common substances.
c (J kg−1 K−1)
Material
human body
3500
methylated spirits
2500
air
1000
aluminium
900
glass
840
iron
440
copper
390
brass
370
lead
130
mercury
140
ice (water)
2100
liquid water
4200
steam (water)
2000
Worked example 11.2.1
CALCULATIONS USING SPECIFIC HEAT CAPACITY
A hot water tank contains 135 L of water. Initially the water is at 20°C.
Calculate the amount of energy that must be transferred to the water to raise the
temperature to 70°C.
Thinking
Working
Calculate the mass of water.
Volume = 135 L
1 L of water = 1 kg
∴ mass of water = 135 kg
ΔT = final temperature − initial
temperature
ΔT = 70 − 20 = 50°C
From Table 11.2.1 the specific
heat capacity of water is
Q = mcΔT
cwater = 4200 J kg−1 K−1.
Use the equation:
Q = mcΔT
= 135 × 4200 × 50
= 28 350 000
= 28 MJ
Worked example: Try yourself 11.2.1
CALCULATIONS USING SPECIFIC HEAT CAPACITY
A bath contains 75 L of water. Initially the water is at 50°C.
Calculate the amount of energy that must be transferred from the water to cool
the bath to 30°C.
CHAPTER 11 | THERMODYNAMICS
309
Worked example 11.2.2
COMPARING SPECIFIC HEAT CAPACITIES
Different states of matter of the same substance have different specific heat
capacities.
What is the ratio of the specific heat capacity of liquid water to that of ice?
Thinking
Working
See Table 11.2.1 for the specific heat
capacities of water in different states.
cwater = 4200 J kg−1 K−1
Divide the specific heat of water by the
specific heat of ice.
Note that ratios have no units since
the unit of each quantity is the same
and cancels out.
cice = 2100 J kg−1 K−1
Ratio =
cwater
cice
=
4200
2100
Ratio = 2
Worked example: Try yourself 11.2.2
COMPARING SPECIFIC HEAT CAPACITIES
What is the ratio of the specific heat capacity of liquid water to that of steam?
PHYSICSFILE ICT CC
Specific heat capacity of water
One of the notable values in the table of specific heat capacities is the high value
for water. It is 10 times, or an order of magnitude, higher than those of most metals
listed. The specific heat capacity of water is higher than those of most common
materials. As a result, water makes a very useful cooling and heat storage agent, and
is used in areas such as generator cooling towers and car-engine radiators.
Life on Earth also depends on the specific heat capacity of water. About 70% of
the Earth’s surface is covered by water, and these water bodies can absorb large
quantities of thermal energy without great changes in temperature. Oceans both
heat up and cool down more slowly than the land areas next to them. This helps to
maintain a relatively stable range of temperatures for life on Earth.
Scientists are now monitoring the temperatures of the deep oceans in order to
determine how the ability of oceans to store large amounts of energy may affect
climate change.
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11.2 Review
SUMMARY
-- ΔT is the change in temperature (in °C or K)
• When heat is transferred to or from a system or
object, the temperature change depends upon
the amount of energy transferred, the mass of the
material(s) and the specific heat capacity of the
material(s): Q = mcΔT,
-- c is the specific heat capacity of the material
(in J kg−1 K−1).
• A substance has different specific heat capacities
when it is in different states (solid, liquid, gas).
where:
-- Q is the heat energy transferred in joules (in J)
-- m is the mass of material being heated in
kilograms (in kg)
KEY QUESTIONS
1
Equal masses of water and aluminium are heated
through the same temperature range. Using the values
of c from Table 11.2.1 on page 309, which material
requires the most energy to achieve this result?
6
Equal amounts of energy are absorbed by equal
masses of aluminium and water. What is the ratio
of the temperature rise of the aluminium to that of
water?
2
Which has the most thermal energy: 10 kg of iron at
20°C or 10 kg of aluminium at 20°C?
7
3
100 mL of water is heated to change its temperature
from 15°C to 20°C. How much energy is transferred to
the water to achieve this temperature change?
4
150 mL of water is heated from 10°C to 50°C. What
amount of energy is required for this temperature
change to occur?
5
For a 1 kg block of aluminium, x J of energy are needed
to raise the temperature by 10°C. How much energy,
in J, is needed to raise the temperature by 20°C?
Which one or more of the following statements about
specific heat capacity is true?
A All materials have the same specific heat capacity
when in solid form.
B The specific heat capacity of a liquid form of a
material is different from that of the solid and
gas forms.
C Good conductors of heat generally have high
specific heat capacities.
D Specific heat capacity is independent of
temperature.
CHAPTER 11 | THERMODYNAMICS
311
11.3 Latent heat
If water is heated, its temperature rises. If enough energy is transferred to the water,
it eventually boils. The water changes state (from liquid to gas). The latent heat
is the energy released or absorbed during a change of state. ‘Latent’ means hidden
or unseen. While a substance changes state, its temperature remains constant. The
energy used in, say, melting ice into water is hidden in the sense that the temperature
does not rise while the change of state is occurring.
ENERGY AND CHANGE OF STATE
Look at the heating curve for water shown in Figure 11.3.1. This graph shows how
the temperature of water changes as energy is added at a constant rate. Although
the rate at which the energy is added is constant, the increase in temperature is not
always constant. There are sections of increasing temperature, and sections where
the temperature remains unchanged (the horizontal sections) while the material
changes state. Temperature remains constant during the change in state from ice to
liquid water and again from liquid water to steam.
Heating curve for water
vapour (steam)
Temperature (°C)
140
120
water boiling
100
80
60
ice melting
liquid water
40
20
0
–20
solid ice
1
2
3
4
5
6
Time (min)
7
8
9
10
FIGURE 11.3.1 A heating curve for water shows that during a change of state, such as ice melting to
water, the temperature remains constant.
Latent heat
The energy needed to change the state of a substance (e.g. solid to liquid, liquid to
gas) is called latent heat. Latent heat is the ‘hidden’ energy that has to be added or
removed from a material in order for the material to change state.
The latent heat is calculated using the equation:
Q = mL
where:
Q is the heat energy transferred in joules (in J)
m is the mass in kilograms (in kg)
L is the latent heat (in J kg−1).
LATENT HEAT OF FUSION (MELTING)
As thermal energy is transferred to a solid, the temperature of the solid increases.
The particles within the solid gain internal energy (as kinetic energy and some
potential energy) and their speed of vibration increases. At the point where the solid
begins to melt, the particles move further apart, reducing the strength of the bonds
holding them in place. At this point, instead of increasing the temperature, the extra
energy increases the potential energy of the particles, reducing the interparticle
or intermolecular forces. No change in temperature occurs, because all the extra
energy supplied is used in reducing these forces between particles.
312
MODULE 3 | WAVES AND THERMODYNAMICS
The amount of energy required to melt a solid is exactly the same as the amount
of potential energy released when the liquid re-forms into a solid. It is termed the
latent heat of fusion.
The amount of energy required depends on the particular solid.
For a given mass of a substance:
heat energy transferred = mass of substance × specific latent heat of fusion
Q = m Lfusion
where:
Q is the heat energy transferred in joules (J)
m is the mass in kilograms (kg)
Lfusion is the latent heat of fusion (J kg−1).
It takes almost 80 times as much energy to turn 1 kg of ice into water (with no
temperature change) as it does to raise the temperature of 1 kg of water by 1°C. It
takes a lot more energy to overcome the large intermolecular forces within the ice
than it does to simply add kinetic energy in raising the temperature.
The latent heats of fusion for some common materials are shown in Table 11.3.1.
TABLE 11.3.1
The latent heat of fusion for some common materials.
Substance
Lfusion (J kg−1)
Melting point (°C)
water
oxygen
lead
ethanol
silver
5
0
3.34 × 10
−219
0.14 × 10
327
0.25 × 10
−114
1.05 × 10
961
0.88 × 10
5
5
5
5
Worked example 11.3.1
LATENT HEAT OF FUSION
How much energy must be removed from 2.5 L of water at 0°C to produce a
block of ice at 0°C? Express your answer in kJ.
Thinking
Working
Cooling from liquid to solid involves the latent
heat of fusion, where the energy is removed
from the water.
1 L of water = 1 kg, so
2.5 L = 2.5 kg
Calculate the mass of water involved.
Use Table 11.3.1 to find the latent heat of
fusion for water.
Lfusion = 3.34 × 105 J kg−1
Use the equation:
Q = m Lfusion
Q = m Lfusion
Convert to kJ.
= 2.5 × 3.34 × 105
= 8.35 × 105 J
Q = 8.35 × 102 kJ
Worked example: Try yourself 11.3.1
LATENT HEAT OF FUSION
How much energy must be removed from 5.5 kg of liquid lead at 327°C to
produce a block of solid lead at 327°C? Express your answer in kJ.
PA
3.6
PA
3.6
CHAPTER 11 | THERMODYNAMICS
313
LATENT HEAT OF VAPORISATION (BOILING)
It takes much more energy to convert a liquid to a gas than it does to convert a solid
to a liquid. This is because, to convert to a gas, the intermolecular bonds must be
broken. During the change of state, the energy supplied is used solely in overcoming
intermolecular bonds. The temperature will not rise until all of the material in the
liquid state is converted to a gas, assuming that the liquid is evenly heated. For
example, when liquid water is heated to boiling point, a large amount of energy is
required to change its state from liquid to steam (gas). The temperature will remain
at 100°C until all of the water has turned into steam. Once the water is completely
converted to steam, the temperature can start to rise again.
The amount of energy required to change a liquid to a gas is exactly the same as
the potential energy released when the gas returns to a liquid. It is called the latent
heat of vaporisation.
The amount of energy required depends on the particular substance.
For a given mass of a substance:
heat energy transferred = mass of substance × specific latent heat of vaporisation
Q = m Lvapour
where:
Q is the heat energy transferred in joules (in J)
m is the mass in kilograms (in kg)
Lvapour is the latent heat of vaporisation (in J kg−1).
Note that, in just about every case, the latent heat of vaporisation of a substance
is different from the latent heat of fusion for that substance. Some latent heat of
vaporisation values are listed in Table 11.3.2.
In many instances, it is necessary to consider the energy required to heat a
substance and also change its state. Problems like this are solved by considering the
rise in temperature separately from the change of state.
TABLE 11.3.2
Substance
The latent heat of vaporisation of some common materials.
Lvapour (J kg−1)
Boiling point (°C)
water
oxygen
lead
ethanol
silver
5
100
22.5 × 10
−183
2.2 × 10
1750
9.0 × 10
78
8.7 × 10
2193
23.0 × 10
5
5
5
5
Worked example 11.3.2
CHANGE IN TEMPERATURE AND STATE
50 mL of water is heated from a room temperature of 20°C to its boiling point at
100°C. It is boiled at this temperature until it is completely evaporated.
How much energy in total was required to raise the temperature and boil
the water?
314
Thinking
Working
Calculate the mass of water involved.
50 mL of water = 0.05 kg
Find the specific heat capacity of water
(see Table 11.2.1 on page 309).
c = 4200 J kg−1 K−1
MODULE 3 | WAVES AND THERMODYNAMICS
Use the equation Q = mcΔT to
calculate the heat energy required to
change the temperature of water from
20°C to 100°C.
Q = mcΔT
Use Table 11.3.2 to find the latent heat
of vaporisation for water.
Lvapour = 22.5 × 105 J kg−1
Use the equation Q = mLvapour to
calculate the latent heat required to
boil water.
Q = mLvapour
Find the total energy required to raise
the temperature and change the state
of the water.
Total Q = 16 800 + 112 500
= 0.05 × 4200 × (100 − 20)
PHYSICSFILE ICT
= 16 800 J
Extinguishing fire
= 0.05 × 22.5 × 105
= 112 500 J
= 129 300 J (or 1.29 × 105 J)
Worked example: Try yourself 11.3.2
CHANGE IN TEMPERATURE AND STATE
The latent heat of vaporisation of
water is very high. This is due to the
molecular structure of the water. This
characteristic of water makes it very
useful for extinguishing fires. That
is because water can absorb vast
amounts of thermal energy before it
evaporates. By pouring water onto
a fire, energy is transferred away
from the fire to heat the water. Then,
even more (in fact much more) heat
is transferred away from the fire to
convert the water into steam.
3 L of water is heated from a fridge temperature of 4°C to its boiling point at
100°C. It is boiled at this temperature until it is completely evaporated.
How much energy in total was required to raise the temperature and boil
the water?
EVAPORATION AND COOLING
If you spill some water on the f loor then come back in a couple of hours, the water
will probably be gone. It will have evaporated. It has changed from a liquid into a
vapour at room temperature in a process called evaporation. The reason for this is
that the water particles, if they have sufficient energy, are able to escape through the
surface of the liquid into the air. Over time, no liquid remains.
Evaporation is more noticeable in volatile liquids such as methylated spirits,
mineral turpentine, perfume and liquid paper. The surface bonds are weaker in these
liquids and they evaporate rapidly. This is why you should never leave the lid off
bottles of these liquids. They are often stored in narrow-necked bottles for this reason.
Whenever evaporation occurs, higher-energy particles escape the surface of the
liquid, leaving the lower-energy particles behind, as is shown in Figure 11.3.2. As a
result, the average kinetic energy of the particles remaining in the liquid drops and
the temperature decreases. Humans use this cooling principle when sweating to stay
cool. When rubbing alcohol is dabbed on your arm before an injection, the cooling
of the volatile liquid numbs your skin.
The rate of evaporation of a
liquid depends on:
• the volatility of the liquid:
more-volatile liquids evaporate
faster
• the surface area: greater
evaporation occurs when
greater surface areas are
exposed to the air
• the temperature: hotter liquids
evaporate faster
• the humidity: less evaporation
occurs in more humid
conditions
• air movement: if a breeze
is blowing over the liquid’s
surface, evaporation is
more rapid.
fast-moving molecules escape
vapour
slower-moving
molecules fall back
evaporation at surface
liquid
slow-moving molecules left behind
FIGURE 11.3.2 Fast-moving molecules with high kinetic energy can escape the liquid, leaving
molecules with lower kinetic energy behind.
CHAPTER 11 | THERMODYNAMICS
315
11.3 Review
SUMMARY
• When a solid material changes state, energy is
needed to separate the particles by overcoming
the attractive forces between the particles.
• The latent heat of vaporisation, Lvapour, is the
energy required to change 1 kg of a material
between the liquid and gaseous states.
• Latent heat is the energy required to change the
state of 1 kg of material at a constant temperature.
• The latent heat of fusion of a material is different
from (and usually less than) the latent heat of
vaporisation for that material.
• In general, for any mass of material the energy
required (or released) is Q = mL.
where:
• Evaporation is when a liquid turns into gas at
room temperature. The temperature of the liquid
falls as this occurs.
Q is the energy transferred in joules (in J)
• The rate of evaporation depends on the volatility,
temperature and surface area of the liquid and the
presence of a breeze.
m is the mass in kilograms (in kg)
−1
L is the latent heat (in J kg ).
• The latent heat of fusion, Lfusion, is the energy
required to change 1 kg of a material between the
solid and liquid states.
KEY QUESTIONS
Refer to the values in Table 11.3.1 and Table 11.3.2. You
may also need to refer to Table 11.2.1 on page 309.
The graph below represents the heating curve for
mercury, a metal that is a liquid at normal room
temperature. Thermal energy is added to 10 g of solid
mercury, initially at a temperature of −39°C until all of
the mercury has evaporated.
Temperature (°C)
1
How much heat energy must be transferred away
from 100 g of steam at 100°C to change it completely
to a liquid?
3
A painter spills some mineral turpentine onto a
concrete floor. After a minute, most of the liquid is
gone. Which of the following is correct?
A Most of the liquid has boiled, becoming hotter.
B Most of the liquid has evaporated and the
remaining liquid becomes warmer as it does so.
C Most of the liquid has evaporated with no change in
temperature of the remaining liquid.
D Most of the liquid has evaporated and the
remaining liquid becomes colder as it does so.
4
A block of ice (200 g) is removed from a freezer and
left on the kitchen bench. How much energy must
be added to melt it to a liquid at room temperature
(20°C)?
357
–39
0
126
670
Energy (J)
3520
a Why does the temperature remain constant during
the first part of the graph?
b What is the melting point of mercury, in degrees
Celsius?
c What is the boiling point of mercury, in degrees
Celsius?
d From the graph, what is the latent heat of fusion of
mercury?
e From the graph, what is the latent heat of
vaporisation of mercury?
316
2
MODULE 3 | WAVES AND THERMODYNAMICS
11.4 Conduction
If two objects are at different temperatures and are in thermal contact (that is,
they can exchange energy via heat processes), then thermal energy transfers from
the hotter object to the cooler object. Figure 11.4.1 shows how, by preventing the
penguin chick’s thermal contact with the cold ice, this adult penguin is able to
protect the vulnerable penguin offspring.
There are three ways that heat can be transferred:
• conduction
• convection
• radiation.
This section focuses on conduction, the following two sections focus on
convection and radiation.
CONDUCTORS AND INSULATORS
Conduction is the process by which heat is transferred from one place to another
without the net movement of particles (atoms or molecules). Conduction can occur
within a material or between materials that are in thermal contact. For example, if one
end of a steel rod is placed in a fire, heat travels along the rod so that the far end of the
rod also heats up; or, if you hold an ice cube, heat travels from your hand to the ice.
While all materials conduct heat to some extent, this process is most significant
in solids. It is important in liquids but plays a lesser role in the movement of energy
in gases.
Materials that conduct heat readily are referred to as good conductors. Materials
that are poor conductors of heat are referred to as insulators. An example of a
good conductor and a good insulator can be seen in Figure 11.4.2.
In secondary physics, the terms ‘conductor’ and ‘insulator’ are used in the
context of both electricity and heating processes. What makes a material a good
conductor of heat does not necessarily make it a good conductor of electricity.
The two types of conduction are related but it is important not to confuse the two
processes. A material’s ability to conduct heat depends on how conduction occurs
within the material.
FIGURE 11.4.1 Emperor penguin chicks avoid
heat loss through conduction by sitting on the
adult’s feet. In this way they avoid contact with
the ice.
ceramic tile
carpet mat
heat retained
heat conducted away
Ceramic floor tiles are good conductors of heat. They conduct heat away from the
foot readily and so your feet feel cold on tiles. The carpet mat is a thermal insulator. Thermal energy
from the foot is not transferred away as quickly and so your foot does not feel as cold.
FIGURE 11.4.2
•
•
Conduction can happen in two ways:
energy transfer through molecular or atomic collisions
energy transfer by free electrons.
CHAPTER 11 | THERMODYNAMICS
317
THERMAL TRANSFER BY COLLISION
direction of
heat flow
FIGURE 11.4.3 Thermal energy is passed on by
collisions between adjacent particles.
The kinetic particle model explains that particles in a solid substance are constantly
vibrating within the material structure and so interact with neighbouring particles.
If one part of the material is heated, then the particles in that region vibrate
more rapidly. Interactions with neighbouring particles pass on this kinetic energy
throughout the system via the bonds between the particles (Figure 11.4.3).
The process can be quite slow since the mass of the particles is relatively large and
the vibrational velocities are fairly low. Materials for which this method of conduction is
the only means of heat transfer are likely to be poor conductors of heat or even thermal
insulators. Materials such as glass, wood and paper are poor conductors of heat.
THERMAL TRANSFER BY FREE ELECTRONS
GO TO ➤ Section 12.1 page 340
Some materials, particularly metals, have electrons that are not directly involved
in any one particular chemical bond. Therefore, these electrons are free to move
throughout the lattice of positive ions.
For example, if a metal is heated, then not only will the positive ions within the
metal gain extra energy but so will these free electrons. As the electron’s mass is
considerably less than the positive ions, even a small energy gain results in a very large
gain in velocity. Consequently, these free electrons provide a means by which heat can
be quickly transferred throughout the whole of the material. It is therefore no surprise
that metals, which are also good electrical conductors because of these free electrons,
are also good thermal conductors. Electrical conductors are discussed in Chapter 12.
THERMAL CONDUCTIVITY
Thermal conductivity describes the ability of a material to conduct heat. It is
temperature dependent and is measured in watts per metre per kelvin ( W m−1 K−1).
Table 11.4.1 highlights the difference in conductivity in metals compared with
other substances.
TABLE 11.4.1
Thermal conductivities of some common materials.
Material
Conductivity (W m−1 K−1)
silver
420
copper
380
aluminium
240
steel
ice
60
2.2
brick, glass
≈1
concrete
≈1
water
0.6
human tissue
0.2
wood
0.15
polystyrene
0.08
paper
0.06
fibreglass
0.04
air
0.025
Factors affecting thermal conduction
The rate at which heat is transferred through a system depends on the:
• nature of the material. The greater a material’s thermal conductivity, the more
rapidly it conducts heat energy.
318
MODULE 3 | WAVES AND THERMODYNAMICS
•
temperature difference between the two objects. Greater temperature difference
results in a faster rate of energy transfer.
• thickness of the material. Thicker materials require a greater number of collisions
between particles or movement of electrons to transfer energy from one side to
the other.
• surface area. Increasing the surface area relative to the volume of a system
increases the number of particles involved in the transfer process, increasing the
rate of conduction.
The rate at which heat is transferred is measured in joules per second (J s−1), or
watts (W).
+ ADDITIONAL
Glazing
Large windows provide light, ventilation and daytime heating. At night or on
cold, cloudy days windows are a major source of heat loss. Glazing (glass) has
a major impact on the energy efficiency of the whole building. On a winter’s
night when it is 15°C colder outside, an average home (with 70 m2 of clear
glass windows, glazed doors and aluminium frames) loses thermal energy at
a rate of approximately 6.5 kW. That’s equivalent to the total heat output of
a large gas heater or an air conditioner running at full capacity. The majority
of that heat loss is by conduction through the glass of the window. For this
reason, windows are a major part of energy-efficient home design.
Insulating glass, usually in the form of double glazing, helps to stop the
transfer of heat by conduction. Double glazing provides resistance to this type
of heat transfer by utilising a sealed space between the panes of glass, as
shown in Figure 11.4.4. The air in the gap between the glass panes is a poor
conductor of heat, which means that much less thermal energy is transferred
through a window that is double glazed.
aluminium
spacer
PHYSICSFILE ICT
Igloos
It seems strange that an igloo can
keep a person warm when ice is so
cold. Igloos are constructed from
compressed snow that contains
many air pockets. The air in these
pockets is a poor conductor of heat,
which means heat inside the igloo is
not easily transferred away. The body
heat of the occupant, as well as that
of their small heat source, is trapped
inside the igloo and able to keep
them warm.
sealed
air space
airtight seal
two panes
of glass
FIGURE 11.4.4 The air
space between the panes
of glass in double-glazed
desiccant to
windows limits thermal
absorb moisture conduction.
Increasingly, argon gas is being used to fill the space between the glass
panes instead of air. As well as having a lower conductivity than air, argon is
also widely available and cheap. Good use of double glazing can reduce heat
loss or gain by 50%, or more when additional treatments are added to reduce
thermal transfer by radiation.
CHAPTER 11 | THERMODYNAMICS
319
THERMAL CONDUCTIVITY
The rate of energy transfer by conduction (energy per unit time) through a material
can be calculated using:
Q
t
= kAd∆T
where:
is the rate of heat energy transferred (in J s−1)
k is the thermal conductivity of the material (in W m−1 K−1)
A is the surface area perpendicular to the direction of heat flow (in m2)
ΔT is the temperature difference across the material in kelvin or degrees
Celsius (in °C or K)
d is the thickness of the material through which the heat is being
transferred (in m).
Q
t
Remember, the unit joules per second ( J s−1) is equivalent to watts (W).
Clothing designers use this relationship when calculating the insulating ability of
parkas and other cold-weather clothes. Architects and builders use it to calculate the
efficiency of building insulation.
Guidelines exist to ensure the efficiency of insulating materials. Building
materials that limit the transfer of heat help to keep houses warm in winter and cool
in summer. This saves money and helps to reduce carbon dioxide emissions from
the use of gas or electricity to heat houses.
Worked example 11.4.1
CALCULATING THE ENERGY TRANSFER BY CONDUCTION THROUGH A
WOODEN WALL
Calculate the rate of energy transfer by conduction through a house wall which
is a 1 m2 section of wood that is 10 cm thick. Assume the inside temperature
in the house is 21°C, and the outside temperature is 10°C. Use Table 11.4.1 on
page 318 to find the thermal conductivity of wood.
Thinking
Write out the equation for the rate of
energy transfer by conduction.
Determine the quantities for the
variables k, A, ΔT and d.
Working
Q
t
=
kA∆T
d
k = 0.15 W m−1 K−1
A = 1 m2
ΔT = 21 − 10 = 11 K
d = 0.1 m
Calculate the rate of energy transfer.
Q
t
=
kA∆T
d
=
0.15 × 1 × 11
0.1
= 16.5 W
320
MODULE 3 | WAVES AND THERMODYNAMICS
Worked example: Try yourself 11.4.1
CALCULATING THE ENERGY TRANSFER BY CONDUCTION THROUGH
A BRICK WALL
Calculate the rate of energy transfer by conduction through a 1 m × 1 m square
section of a brick wall. Assume the wall is 5 cm thick, the inside temperature is
21°C, and the outside temperature is −4°C. Use Table 11.4.1 on page 318 to find
the thermal conductivity of brick.
11.4 Review
SUMMARY
• Conduction is the process of heat transfer within a
material or between materials without the overall
transfer of the substance itself.
-- Heat transfer by molecular collisions and
free electrons occurs in good to very good
conductors.
• All materials conduct heat to a greater or lesser
degree. Materials that readily conduct heat are
called good thermal conductors. Materials that
conduct heat poorly are called thermal insulators.
• The rate of conduction depends on the
temperature difference between two materials, the
thickness of the material, the surface area and the
nature of the material.
• Whether a material is a good conductor depends
on the method of conduction:
• The rate of conduction can be calculated using
Q
= kAd∆T , where k is the thermal conductivity, A is
t
the surface area, ΔT is the change in temperature
and d is the thickness of the material.
-- Heat transfer by molecular collisions alone
occurs in poor to very poor conductors.
KEY QUESTIONS
1
Explain why the process of conduction by molecular
collision is slow.
2
Why are metals more likely to conduct heat than
wood?
3
List the properties of a material that affect its ability to
conduct heat.
4
Stainless steel saucepans are often manufactured with
a copper base. What is the most likely reason for this?
5
One way of making a house energy-efficient is to use
double-glazed windows. These consist of two panes
of glass with air trapped between the panes. On a hot
day, the energy from the hot air outside the house is
not able to penetrate the air gap and so the house
stays cool. Which of the following best explains why
double-glazing works?
A Air is a conductor of heat and so the thermal
energy is able to pass through.
B Air is a conductor of heat and so the thermal
energy is not able to pass through.
C Air is an insulator of heat and so the thermal
energy is not able to pass through.
D Air is an insulator of heat and so the thermal
energy is able to pass through.
6
Fibreglass insulation batts are thick and lightweight,
and they make a house more energy-efficient. On a
cold July night, the external temperature in the roof of
an insulated house is 6°C. The air temperature near
the ceiling inside the house is 20°C.
Complete the paragraph below by choosing the
correct response from the choices given to explain the
effects of ceiling insulation in this situation.
The insulation batts stop the thermal energy from
escaping/entering the house. The air in the batts has
high/low conductivity and the thermal energy is able/
not able to escape from the house.
7
On a cold day, the plastic or rubber handles of a
bicycle feel much warmer than the metal surfaces.
Explain this in terms of the thermal conductivity of
each material.
CHAPTER 11 | THERMODYNAMICS
321
11.5 Convection
This section investigates convection, which is the transfer of thermal energy within a
f luid (liquid or gas) by the movement of hot areas from one place to another. Unlike other
forms of heat transfer such as conduction and radiation, convection involves the mass
movement of particles within a system over a distance that can be quite considerable.
HEATING BY CONVECTION
convection
current
bunsen burner
FIGURE 11.5.1 When a liquid or gas is heated, it
becomes hotter and less dense, so it rises. The
colder, denser fluid falls. As this fluid heats up, it
in turn rises, creating a convection current.
Although liquids and gases are generally not very good conductors of thermal energy,
heat can be transferred quite quickly through liquids and gases by convection. Unlike
other forms of thermal energy transfer, convection involves the mass movement of
particles within a system over a distance.
As a f luid is heated, the particles within it gain kinetic energy and push apart due
to the increased vibration of the particles. This causes the density of the heated f luid
to decrease and the heated f luid rises. Colder f luid, with slower-moving particles, is
more dense and heavier and hence falls, moving in to take the place of the warmer
f luid. A convection current forms when there is warm f luid rising and cool f luid
falling. This action can be seen in Figure 11.5.1. Upwellings in oceans, wind and
weather patterns are at least partially due to convection on a very large scale.
It is difficult to quantify the thermal energy transferred via convection but some
estimates can be made. The rate at which convection occurs is affected by:
• the temperature difference between the heat source and the convective f luid
• the surface area exposed to the convective f luid.
In a container, the effectiveness of convection to transfer heat depends on the
placement of the source of heat. For example, the heating element in a kettle is
always found near the bottom of the kettle. From this position, convection currents
form throughout the water to heat it more effectively as seen in Figure 11.5.2a. If
the heating element is placed near the top of the kettle, convection currents form
only near the top. This is because the hotter water is less dense than the cooler water
below and stays near the top. In Figure 11.5.2b convection currents do not form
throughout the water.
(a)
(b)
100°C
100°C
100°C
87°C
100°C
45°C
WS
3.9
FIGURE 11.5.2 (a) By placing the heating element at the bottom of a kettle, the water near the
bottom is heated and rises, forming convection currents throughout the entire depth of the water.
(b) If the heating element is placed near the top of the kettle, the convection currents form near the
top and heat transfer is slower.
There are two main causes of convection:
forced convection, for example ducted heating in which air is heated and then
blown into a room
• natural convection, such as that illustrated in Figure 11.5.1, when a f luid rises
as it is heated.
A dramatic example of natural convection is the thunderhead clouds of summer
storms (Figure 11.5.3), which form when hot, humid air from natural convection
currents is carried rapidly upwards into the cooler upper atmosphere.
•
FIGURE 11.5.3 The thunderheads of summer
storms are a very visible indication of natural
convection in action.
322
MODULE 3 | WAVES AND THERMODYNAMICS
PHYSICS IN ACTION
ICT
Wind chill
Convective effects are the main means of heat transfer that lead to the ‘wind
chill’ factor. The wind blows away the thin layer of relatively still air near
the skin that would normally act as a partial insulator in still air. Cooler air
comes in closer contact with the skin and heat loss increases. It feels as if
the ‘effective’ temperature of the surrounding air has decreased. Skiers can
experience similar effects simply from the wind created by their own motion.
In cold climates the wind chill factor can become an important factor
to consider. The chilling effect is even more dramatic when the body or
clothing is wet, increasing evaporative cooling. Bushwalkers look for clothing
that dries rapidly after rain and which carries moisture from the perspiration
of heavy exertion away from the skin.
PHYSICSFILE ICT
Paragliders
Paragliders fly by sitting in a harness suspended beneath a fabric wing
(Figure 11.5.4). They gain altitude by catching thermals. Thermals are columns of
rising hot air created by dark regions on the ground that have been heated up by the
Sun. Roads, rock faces and ploughed fields are good at creating thermals.
In 2007, Polish paraglider Ewa Wisnierska was practising in NSW for a competition
when she was caught in an intense thermal updraught during a storm. She reached
an altitude of almost 10 km. Fortunately she lost altitude and landed about 60 km
from where she started, where her crew found and rescued her.
FIGURE 11.5.4 Paragliders can gain altitude by finding a thermal. These are areas of rising
hot air created by hot regions on the ground.
CHAPTER 11 | THERMODYNAMICS
323
11.5 Review
SUMMARY
• Convection is the transfer of heat within a fluid
(liquid or gas).
• A convection current forms when there is warm
fluid rising and cool fluid falling.
• Convection involves the mass movement of
particles within a system over a distance.
KEY QUESTIONS
1
Through what states of matter can convection occur?
2
In which direction does the transfer of heat in a
convection current initially flow?
3
Pilots of glider aircraft or hang-gliders, some birds
such as eagles and some insects rely on ‘thermals’ to
give them extra lift. Explain how these rising columns
of air are established.
4
324
Which of the following statements about liquids and
gases is true?
A Liquids and gases can transfer heat quite slowly
through convection even though liquids and gases
are very good conductors of thermal energy.
B Liquids and gases can transfer heat quite slowly
through conduction because liquids and gases are
very good conductors of thermal energy.
MODULE 3 | WAVES AND THERMODYNAMICS
C Liquids and gases can transfer heat quite quickly
through conduction because liquids and gases are
very good conductors of thermal energy.
D Liquids and gases can transfer heat quite quickly
through convection even though liquids and gases
are not very good conductors of thermal energy.
5
Convection is referred to as a method of heat transfer
through fluids. Explain whether it is possible for solids
to pass on their heat energy by convection.
6
On a hot day, the top layer of water in a swimming
pool can heat up while the lower, deeper parts of
the water can remain quite cold. Explain, using the
concept of convection, why this happens.
11.6 Radiation
Both convection and conduction involve the transfer of heat through matter. Life on
Earth depends upon the transfer of energy from the Sun through the near-vacuum
of space. If heat could only be transferred by the action of particles, then the Sun’s
energy would never reach Earth. Radiation is a means of transfer of heat without
the movement of matter.
RADIATION
In this context, ‘radiation’ means electromagnetic radiation, which includes visible,
ultraviolet and infrared light. Together with other frequencies of radiation, these
make up the electromagnetic spectrum.
The transfer of heat from one place to another without the movement of particles
is by electromagnetic radiation. Electromagnetic radiation travels at the speed of
light. When electromagnetic radiation (light) hits an object, it is partially ref lected,
partially transmitted and partially absorbed. The absorbed part transfers thermal
energy to the absorbing object and causes a rise in temperature. When you hold a
marshmallow by an open fire, you are using radiation to toast the marshmallow, as
shown in Figure 11.6.1.
FIGURE 11.6.1
Heat transfer from the flame to the marshmallow is an example of radiation.
Electromagnetic radiation is emitted by all objects that are at a temperature
above absolute zero (0 K or −273°C). The wavelength and frequency of the emitted
radiation depends on the internal energy of the object. The higher the temperature of
the object, the higher the frequency and the shorter the wavelength of the radiation
emitted. This can be seen in Figure 11.6.2.
Energy
high temperature
low temperature
High frequency
Low frequency
A system emits radiation over a range of frequencies. At a low temperature, it emits
small amounts of radiation of longer wavelengths. As the temperature of the system increases, more
short-wavelength radiation is emitted and the total radiant energy emitted increases.
FIGURE 11.6.2
CHAPTER 11 | THERMODYNAMICS
325
A human body emits radiation in the infrared range of wavelengths, while hotter
objects emit radiation of a higher frequency and shorter wavelength. Hotter objects
can emit radiation in the range of visible, ultraviolet and shorter wavelengths of the
electromagnetic spectrum. For example, as a red-hot fire poker heats up further, it
becomes yellow-hot.
EMISSION AND ABSORPTION OF RADIANT ENERGY
All objects both absorb and emit thermal energy by radiation. If an object absorbs
more thermal energy than it emits, its temperature increases. If an object emits more
energy than it absorbs, its temperature falls. If no temperature change occurs, the
object and its surroundings are in thermal equilibrium.
While all objects emit some radiation, they don’t all emit or absorb at the
same rate.
A number of factors affect both the rate of emission and the rate of absorption.
• Surface area: The larger the exposed surface area, the higher the rate of
radiant transfer.
• Temperature: The greater the difference between the temperature of the
absorbing or emitting surface and the temperature of the surrounding objects,
the greater the rate of energy transfer by radiation.
• Wavelength of the incident radiation: Matt black surfaces are almost perfect
absorbers of radiant energy at all wavelengths. Highly ref lective surfaces are
good ref lectors of all wavelengths. An example of how ref lective surfaces can
be exploited is shown in Figure 11.6.3. For all other surfaces, the absorption
of particular wavelengths of radiant energy is affected by the wavelength of
that energy. For example, although white surfaces absorb visible wavelengths
of radiant energy poorly, white surfaces absorb infrared radiation just as well as
black surfaces do.
• Surface colour and texture: The characteristics of the surface itself determine
how readily that particular surface emits or absorbs radiant energy. Matt black
surfaces absorb and emit radiant energy faster than shiny, white surfaces. This
means that a roughened, dark surface heats up faster than a shiny, light one. Matt
black objects also cool down faster since they radiate energy just as efficiently
as they absorb it. Car radiators are painted black to increase the emission of
thermal energy that is collected from the car engine.
FIGURE 11.6.3 The silvered surface of an
emergency blanket reflects thermal energy back
to the body, and retains the radiant energy,
which would normally be lost. This simple
method works as excellent thermal insulation.
PHYSICSFILE ICT
Thermal imaging
All objects emit radiant energy. Humans are warm-blooded and emit radiation in the
infrared region of the electromagnetic spectrum. This radiation is not visible to us, but
can be detected using thermal imaging devices or night-vision goggles. These devices
are used by the military for surveillance, and by search and rescue personnel. Some
animals, notably some varieties of bugs, are able to detect infrared radiation.
(a)
WS
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326
(b)
FIGURE 11.6.4 (a) The person standing behind the bush is difficult to see in normal light, but
(b) they are warmer than the vegetation surrounding them and produce a stronger infrared
emission, which a thermal imaging device is able to detect.
MODULE 3 | WAVES AND THERMODYNAMICS
11.6 Review
SUMMARY
• Any object whose temperature is greater than
absolute zero emits thermal energy by radiation.
• The rate of emission or absorption of radiant heat
depends upon the:
• Radiant transfer of thermal energy from one place
to another occurs by means of electromagnetic
waves.
-- temperature difference between the object and
the surrounding environment
-- surface area and surface characteristics of the
object
• When electromagnetic radiation falls on an object,
it is partially reflected, partially transmitted and
partially absorbed.
-- wavelength of the radiation.
KEY QUESTIONS
1
2
Light is shone on an object.
a List three interactions that can occur between the
light and the object.
b Which of the interactions from part (a) are
associated with a rise in temperature?
The wavelength and frequency of emitted radiation
depends on the internal energy of an object. Complete
the sentences below by choosing the correct option
from those provided.
The higher the temperature of the object, the higher/
lower the frequency and the longer/shorter the
wavelength of the radiation emitted. For example,
if a particular object emits radiation in the visible
range, a cooler object could emit light in the infrared/
ultraviolet range of the electromagnetic spectrum.
3
4
Why is it impossible for heat to travel from the Sun to
the Earth by conduction or convection?
5
Three identical, sealed beakers are filled with nearboiling water. One beaker is painted matt black, one is
dull white and the third is gloss white.
a Which beaker will cool fastest?
b Which beaker will cool slowest?
6
Computer chips generate a lot of thermal energy
that must be dispersed for the computer to function
efficiently. Devices called heat sinks are used to help
this process. What would you predict the heat sinks to
be made of?
Which of the following will affect the rate at which an
object radiates thermal energy?
A its temperature
B its colour
C its surface nature (shiny or dull)
D none of these
E all of A–C.
CHAPTER 11 | THERMODYNAMICS
327
Chapter review
KEY TERMS
absolute zero
acceleration
collinear
absorption
displacement
conduction
distance travelled
conductor
position
convection
electromagnetic spectrum
emit
KEY QUESTIONS
evaporation
gas
heat
scalar
insulator
speed
internal
vector energy
kelvin
vector diagram
kinetic
velocityparticle model
latent heat
latent heat of fusion
latent heat of vaporisation
liquid
radiation
solid
specific heat capacity
temperature
thermal contact
thermal energy
thermal equilibrium
volatile
REVIEW QUESTIONS
1
According to the kinetic particle model, which of the
following can be said of particles of matter?
A They are in constant motion.
B They have different sizes.
C They have different shapes.
D They are always floating.
2
What does the kelvin scale measure?
3
How does temperature differ from heat?
4
Convert:
a 5°C to kelvin
b 200 K to °C.
5
A chef vigorously stirs a pot of cold water and does
150 J of work on the water. The water also gains 75 J of
thermal energy from the surroundings. Calculate the
change in energy of the water.
6
A scientist very carefully does mechanical work on a
container of liquid sodium. The liquid sodium loses
300 J of energy to its surroundings but gains 250 J of
energy overall. How much work did the scientist do?
7
Tank A is filled with hydrogen gas at 0°C and another
tank, B, is filled with hydrogen gas at 300 K. Describe
the difference in the average kinetic energy of the
hydrogen particles in each tank.
8
The specific heat capacity of iron is approximately
half that of aluminium. A ball of iron and a ball of
aluminium, both at 80°C, are dropped into a thermally
insulated jar that contains a mass of water, equal
to that of the balls, at 20°C. Thermal equilibrium is
eventually reached. Describe the final temperatures of
each of the metal balls.
9
If 4.0 kJ of energy is required to raise the temperature
of 1.0 kg of paraffin by 2.0°C, how much energy (in kJ)
is required to raise the temperature of 5.0 kg of
paraffin by 1.0°C?
10 A cup holds 250 mL of water at 20°C. 10.5 kJ of heat
energy is transferred to the water. What temperature
does the water reach after the heat is transferred?
328
MODULE 3 | WAVES AND THERMODYNAMICS
11 A block of iron is left to cool. After cooling for a short
time, 13.2 kJ of energy has been transferred away from
the block of iron and its temperature has decreased by
30°C. What is the mass of the block of iron?
12 Two cubes, one silver and one iron, have the same
mass and temperature. A quantity Q of heat is
removed from each cube. Which one of the following
properties causes the final temperatures of the cubes
to be different?
A density
B specific heat capacity
C latent heat of vaporisation
D volume
13 A solid substance is heated but its temperature does
not change. Explain what is occurring.
14 Which possesses the greater internal energy—1 kg
of water boiling at 100°C or 1 kg of steam at 100°C?
Explain why.
15 A liquid is evaporating, causing the liquid to cool.
Explain why the temperature of the liquid decreases.
16 A 2.00 kg metal object requires 5.02 × 103 J of heat
to raise its temperature from 20.0°C to 40.0°C. What
is the specific heat capacity of the metal in J kg −1 K−1?
Give your answer to the nearest whole number.
17 How many joules of energy are required to melt
exactly 80 g of silver? (Lfusion = 0.88 × 105 J kg −1.) Give
your answer to two significant figures.
18 An insulated container holding 4.55 kg of ice at 0.00°C
has 2.65 MJ of work done on it, while a heater provides
14 600 J of heat to the ice. If the latent heat of fusion of
ice is 3.34 × 105 J kg−1, calculate the final temperature
of the water. Assume that the increase in internal
energy is first due to an increase in the potential
energy and then an increase in the kinetic energy.
19 How many kJ of energy are required to melt exactly
100 g of ice initially at −4.00°C? Assume no loss of
energy to surroundings.
20 Which of the following explains why hot water in a spapool evaporates more rapidly than cold water?
A Hot water molecules have less energy than cold
water molecules.
B Hot water molecules have more energy than cold
water molecules.
C Hot water forms water vapour and bubbles to the
surface.
D Hot water dissolves into the pool material more
rapidly.
21 An ice-cube that is sitting on a block of polystyrene
at room temperature (20°C) melts very slowly.
An ice-cube sitting on a metal pan also at room
temperature melts very rapidly. Explain why this is so
and why the metal feels colder to touch, even though
it is the same temperature as the polystyrene.
25 Thermal imaging technology can be used to locate
people lost in the Australian bush. How can thermal
imaging technology ‘see’ people when the naked eye
cannot?
26 Hypothermia is the cooling of the body to levels
considerably lower than normal. The body’s functions
slow and death can result. A person may survive
12 hours in air at 0°C before suffering hypothermia,
but may survive only a few minutes in water at 0°C.
Explain why this is so and how wetsuits can help you
survive in cold water.
27 After completing the activity on page 300, reflect on
the inquiry question: How are temperature, thermal
energy and particle motion related? In your response,
discuss how energy can be transferred in relation to
the model.
22 How does a down-filled quilt keep a person warm
in winter?
23 On a hot day when you sweat, your body feels cooler
when a breeze is blowing. Explain why this happens.
24 A vacuum flask has a tight-fitting stopper at the top.
Its walls are made up of an inner and outer layer,
which are shiny and are separated by a gap from
which most of the air has been removed. Describe
how this design makes a vacuum flask good at
keeping liquids hot inside.
CHAPTER 11 | THERMODYNAMICS
329
MODULE 3 • REVIEW
REVIEW QUESTIONS
MR
3
Waves and thermodynamics
Multiple choice
1
2
3
4
5
330
Three metals A, B and C are placed in thermal contact
with one another. Heat flows from A to B and from C to B.
What can you say about the relative temperatures of
metals A and C?
A A is at a higher temperature than C.
B A is at a lower temperature than C.
C A is at the same temperature as C.
D There is insufficient information to compare the
temperatures of A and C.
Two objects A and B are at the same temperature.
What does this imply about the kinetic energy of the
molecules in each object?
A The kinetic energy of any molecule in object A will be
the same as that of any molecule in object B.
B The average kinetic energy of the molecules in object
A will be the same as in object B.
C The highest kinetic energy of any molecule in object A
will be the same as in object B.
D The lowest kinetic energy of any molecule in object A
will be the same as in object B.
A bucket is filled with equal amounts of hot and cold
water. The hot water is initially at 80°C and the cold
water is initially at 10°C.
(Hint: For all mixtures, heat is transferred from a hotter
body to a cooler one until thermal equilibrium is
reached. So here, mcΔThot water = mcΔTcold water.)
What will be the approximate temperature of the final
mixture?
A 10°CB 45°C
C 70°CD 90°C
Which of the following examples supports the statement
that it takes a larger amount of thermal energy to melt
ice than to warm air?
A Moisture forms on the outside of a glass of ice water.
B Ice cubes in a freezer can be colder than 0°C.
C Snow lasts all summer in some areas of New Zealand.
D Snowstorms can occur at low altitudes in winter.
A chemical engineer is doing a gas law calculation and
understands that she needs to use the kelvin unit for
temperature. Her thermometer in the reactor vessel
shows the temperature as 1550°C.
What is this temperature in kelvin?
A 1277 K
B 1550 K
C 1823 K
D 2732 K
MODULE 3 | WAVES AND THERMODYNAMICS
WS
3.10
WS
3.11
6
Boiling water results in steam formation at the same
temperature. Which one or more of the following has
changed for the water molecules in the steam?
A the average kinetic energy of the particles
B the average potential energy of the particles
C the total internal energy of the particles
D the total number of particles
7
A 150 g block of silver is heated until it starts to melt.
How much energy is required to change the silver into a
liquid?
A 13 kJ
B 1.3 MJ
C 132 kJ
D 132 MJ
8
The Earth is approximately 150 × 106 km from the Sun,
and Mars is a further 78 × 106 km from Earth. How
much is the intensity of light from the Sun reduced at
Mars compared to Earth?
A The intensity of light at Mars is 0.4 times the intensity
at Earth.
B The intensity of light at Mars is 0.2 times the intensity
at Earth.
C The intensity of light at Mars is 2.3 times the intensity
at Earth.
D The intensity of light at Mars is 0.3 times the intensity
at Earth.
9
Which one or more of the following is true for both
mechanical and electromagnetic waves?
A They are light waves.
B They are sound waves.
C They need a medium to travel in.
D They transfer energy.
10 Sound waves of frequency f are diffracted as they pass
through a narrow slit of width w. Which one or more of
the following will increase the diffraction?
A an increase in f
B an increase in w
C a decrease in f
D a decrease in w
11 Which one or more of the following phenomena can be
modelled by a pure wave model of light?
A refraction
B superposition
C reflection
D diffraction
12 Which one or more answers completes the following
statement correctly?
Waves in a violin string will travel faster if the string is:
A lightweight
B heavy
C tight
D loose
13 What is the critical angle for light travelling from glass
(n = 1.5) into air (n = 1.0)?
A 90.0°B 41.8°
C 48.2°
D 45.0°
14 Which one or more of the following properties of a wave
can change when it is reflected?
A period
B frequency
C amplitude
D wavelength
15 In which of the following diagrams are the incident and
reflected angles correct?
B
A
n1
n2
n1
r
i
n2
r
i
D
r
n1
n2
i
n1
n2
19 The frequency of a note played by a flute is 280 Hz.
What is the wave number of the sound wave? (Assume
the speed of sound in air is 340 m s−1.)
B 4.2 m−1
C 5.2 m−1
D 6.2 m−1
A 3.2 m−1
20 When does resonance occur in an object?
A When the driving frequency and the natural frequency
are both equal to zero.
B When the driving frequency is equal to the natural
frequency.
C When the driving frequency is greater than the
natural frequency.
D When the driving frequency is less than the natural
frequency.
Short answer
21 Explain the change in internal energy that occurs when
dry ice turns into gaseous carbon dioxide.
C
18 How would the intensity of sound change if you moved
from a distance r from the source to a distance of 3r?
A It would increase by a factor of 9.
B It would decrease by a factor of 9.
C It would increase by a factor of 3.
D It would decrease by a factor of 3.
r
i
16 What is the first harmonic for an open-ended pipe that
is 30 cm long when the speed of sound in the tube is
140 m s−1?
A 23 Hz
B 58 Hz
C 117 Hz
D 233 Hz
17 Describe the image of an object produced by a convex
mirror when the object is 30 cm away, the focal length
of the mirror is 20 cm, and the radius of curvature of the
mirror is 40 cm.
A real, diminished
B real, enlarged
C virtual, diminished
D virtual, enlarged
22 A student attempts to identify a metal by measuring
its specific heat capacity. A 100 g block of the metal is
heated to 75°C and then transferred to a 70 g copper
calorimeter containing 200 g of water at 20°C. The
temperature of the final mixture is 25°C.
Using the table below, what metal is the student
probably testing?
(Hint: For all mixtures, heat is transferred from a hotter
body to a cooler one until thermal equilibrium is
reached. So here, mcΔThot water = mcΔTcold water.)
Material
Specific heat capacity (J kg−1 K−1)
lead
130
mercury
140
methylated spirits
250
brass
370
copper
390
iron
440
glass
840
aluminium
900
air
1000
steam
2000
ice
2100
human body
3500
liquid water
4200
REVIEW QUESTIONS
331
MODULE 3 • REVIEW
23 Explain why the unit of specific heat capacity is J kg−1 K−1
but the unit for heat of fusion is J kg−1.
24 Calculate the amplitude, period and frequency of the
curve in the graph below.
2
Pressure
1
1
2
3
4
Time (ms)
5
6
7
–1
–2
25 Explain how to create an image of an object using a
concave mirror if the object is a long way from the mirror.
26 Wei is standing on a platform when an approaching
XPT train sounds its horn. The frequency of the horn is
500 Hz and the train is travelling at 126 km h−1. What is
the frequency of the horn as heard by Wei? (Assume the
speed of sound in air is 340 m s−1.)
27 Zanni and Justine are tuning their violins before a
concert and they notice a beat frequency of 3 Hz. Zanni
knows that her violin is playing a concert A at 440 Hz.
What frequencies could Justine be playing, and explain
how they would go about becoming in tune with each
other.
28 Draw a ray diagram for an object 15 cm high placed 40 cm
from a concave lens. The lens has a focal length of 20 cm.
29 A pulse is sent along a 7 m length of rope at t = 0 with a
speed of 1.2 m s−1. A second pulse is sent down the rope
after 1 s. Calculate the position where the two pulses
interact.
30 Keely is standing 50 cm from a concave mirror that has
a focal length of 15 cm and a 30 cm radius of curvature.
a Calculate the position of her image. Is it real or virtual?
b What is the magnification of her image? Is it inverted
or upright?
Extended response
31 a While camping during the holidays, a physics student
notices that the night temperature is much lower
on clear nights than on cloudy nights. Explain this
difference.
b This same student is trying to calculate the energy
lost through their tent walls over a 5 hour period
when the average outdoor temperature is −3°C. What
assumptions are needed to complete this calculation?
332
MODULE 3 | WAVES AND THERMODYNAMICS
c Make some reasonable assumptions for the values
outlined in part b to calculate the energy Q lost in
5 hours. Explain why the temperature may not drop
within the tent while the hiker is inside even though
energy is lost through the tent walls.
32 A ray of light passes through a diamond. On striking
the inside wall of the diamond the ray makes an angle
of 67.0° with the diamond–air boundary. The index of
refraction of the diamond is 2.4. Calculate the:
a angle of incidence
b angle of refraction of the transmitted ray (assuming
nair = 1.0)
c angle of deviation (i.e. the difference between the
angles of incidence and refraction)
d speed of light in the diamond.
33 It is possible to make home-made ice cream by using a
salt-water solution (brine) as a refrigerant for the cream
and sugar mix. You can assume for the purposes of this
exercise that the cream and sugar mix is 70% water. It is
actually the water in the cream that freezes.
Salt is added to ice to make 5.00 kg of cold brine
at−11°C.
500 g of cream and sugar mixture is cooled in a plastic
bag which is plunged into the refrigerant.
Use the following data where appropriate:
Heat of fusion of water: 334 kJ kg−1
Specific heat capacity of water: 4200 J kg−1 K−1
Specific heat capacity of cream and sugar mix:
3.80 × 103 J kg−1 K−1
Specific heat capacity of brine solution: 3.5 × 103 J kg−1 K−1
a Calculate the amount of heat that needs to be
extracted from 500 g of ice cream mixture at 25.0°C
to reduce the temperature of the mixture to 0°C.
b Explain why the specific heat capacity of the mixture
is less than that of water.
c Once the mixture starts to form ice crystals, the
temperature does not drop further. Explain why this is
the case.
d Calculate the heat that would need to be extracted
from the mixture at 0°C to completely freeze the 70%
of water in the mixture.
e Calculate the final temperature of the mixture when
all the water in the ice cream is frozen, assuming no
other heat loss.
34 a The speed of sound in Coco the Clown’s 30.0 cm flute
is 340 m s−1. Calculate the fundamental frequency of
his flute.
b Coco is playing his flute while riding a unicycle at the
circus. He is travelling towards a spectator at 5 m s−1
while playing the flute’s fundamental frequency. What
frequency does the spectator hear? (Assume the
speed of sound in air is 340 m s−1.) Would this be a
noticeable difference?
c Describe how the Doppler effect changes the
observed frequency of sound waves in motion.
d The low C on a flute should have a frequency of
262 Hz. When Jeremy and Lucia both play a low C
on their flutes, they notice a beat frequency of 2 Hz.
If Lucia’s flute is correctly tuned, what frequencies
could Jeremy be playing?
35 a Explain how resonance occurs in a system.
b Explain what could happen if resonance is not taken
into account when designing bridges and buildings.
c A bridge made of rope is suspended across a
gorge. The rope is 100 m long and is tied at both
ends. Calculate the fundamental frequency of the
system in terms of a velocity v. What would be the
corresponding wavelength?
d A large group of people march across the bridge in
step, causing the bridge to vibrate. Describe what
would happen if this causes the bridge to vibrate at
its natural frequency.
REVIEW QUESTIONS
333
MODULE
Electricity and magnetism
Atomic theory and the laws of conservation of energy and electric charge are
unifying concepts in understanding the electrical and magnetic properties and
behaviour of matter. Interactions resulting from these properties and behaviour
can be understood and analysed in terms of electric fields represented by
lines. These representations and mathematical models can be used to make
predictions about the behaviour of objects, and explore the limitations of
the models.
This module also examines how the analysis of the behaviour of electrical
circuits and the transfer and conversion of energy in electrical circuits has led
to a variety of technological applications.
Outcomes
By the end of this module you will be able to:
• develop and evaluate questions and hypotheses for scientific investigation
PH11-1
• analyse and evaluate primary and secondary data and information PH11-5
• communicate scientific understanding using suitable language and
terminology for a specific audience or purpose PH11-7
• explain and quantitatively analyse electric fields, circuitry and
magnetism PH11-11
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
CHAPTER
Electrostatics
Every object around you is made up of charged particles. When these particles
move relative to one another, we experience a phenomenon known as ‘electricity’.
In this chapter you will investigate the concept of electric charge, electric fields, the
effect on point charges within electric fields and how charged objects interact with
other objects (both charged and neutral).
Content
INQUIRY QUESTION
How do charged objects interact with other charged objects
and with neutral objects?
By the end of this chapter you will be able to:
• conduct investigations to describe and analyse qualitatively and quantitatively:
CCT ICT
-- processes by which objects become electrically charged (ACSPH002)
-- the forces produced by other objects as a result of their interactions with
charged objects (ACSPH103)
-- variables that affect electrostatic forces between those objects (ACSPH103)
• using the electric field lines representation, model qualitatively the direction and
strength of electric fields produced by:
-- simple point charges
-- pairs of charges
-- dipoles
-- parallel charged plates ICT
• apply the electric field model to account for and quantitatively analyse interactions
between charged objects using:
-- E = Fq (ACSPH103, ACSPH104)
-- E = − V
-- F =
d
1 q1q2
4 πε 0 r 2
(ACSPH102)
• analyse the effects of a moving charge in an electric field, in order to relate
potential energy, work and equipotential lines, by applying: (ACSPH105)
-- V =
∆U ,
q
where U is potential energy and q is the charge.
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
12.1 Electric charge
PHYSICS INQUIRY CCT
Charged balloons
How do charged objects interact with other charged objects and with neutral objects?
COLLECT THIS…
•
4 water balloons, inflated with air
•
2 normal balloons, one inflated to be large and one small
•
clean, dry vertical surfaces, each made of a different material (metal, wood, plastic, glass)
•
clean hair or rabbit fur
•
cling film or Teflon
•
loose confetti or other light material.
DO THIS…
1 Try to place the water balloons on the different vertical surfaces. What happens?
2 Rub two of the water balloons on clean hair, or rabbit fur. This will charge the balloons with a negative charge.
Observe how the hair/fur behaves near the balloon.
3 Once the balloons have been negatively charged, place on different vertical surfaces. The vertical surfaces are neutral
in charge (neither positive nor negative).
4 Rub the other two balloons on the cling film or Teflon. This will charge the balloons with a positive charge. Observe
how the cling film behaves near the balloon.
5 Repeat step 3 with the positively charged balloons.
6 Recharge the balloons. Place two negatively (or positively) charged balloons next to each other but not touching.
Observe what happens when you let go.
7 Rub both the small and large balloons on the clean hair or rabbit fur for the same amount of time. This will add
approximately the same negative charge to each balloon.
8 Hold the two balloons over the loose confetti. Which balloon attracted more confetti?
RECORD THIS…
Describe how the charge on the balloon changed how it interacted with each object.
Present your results as a poster summarising the forces that electric charges create.
REFLECT ON THIS…
How do charged objects interact with other charged objects and with neutral objects?
When have you seen electrostatic charges interacting before?
electron
All matter in the universe is made of tiny particles. These particles have a property
called charge that can be positive, negative or neutral. Usually, the numbers of
positive and negative charges balance out so perfectly that we are completely
unaware of them. However, when significant numbers of these charged particles
become separated or move relative to each other, it results in electricity.
neutron
nucleus
FIGURE 12.1.1
338
proton
EXISTENCE OF CHARGE CARRIERS
The tiny particles that make up all matter are called atoms. Every atom contains
a nucleus at its centre. A nucleus is made up of positively charged particles called
protons and neutral particles called neutrons. The nucleus, which is positively
charged due to the protons, is surrounded by negatively charged electrons. A
model of an atom is shown in Figure 12.1.1.
A simple model of an atom.
MODULE 4 | ELECTRICITY AND MAGNETISM
Simple models of the atom, often called planetary models, show the electrons
as orbiting the nucleus much like the planets orbit the Sun, as you can see from the
figure. This is because particles with like charges repel each other, but particles with
opposite charges attract each other. In an atom, the negatively charged electrons are
attracted to the positively charged nucleus.
This is an important rule to remember when thinking about the interaction of
charged particles.
In neutral atoms, the number of electrons is exactly the same as the number of
protons. This means that their charges balance each other out, leaving the atom
electrically neutral.
It is difficult to remove a proton from the nucleus of an atom. In comparison,
electrons are loosely held to their respective atoms and it is relatively easy for them
to be removed.
When electrons move from one object to another, each object is said to have
gained a net charge. The object that loses the electrons will gain a net positive
charge, since it will now have more positive protons than negative electrons. The
object that gains electrons will gain a net negative charge. When an atom has gained
or lost electrons, we say it has been ionised or has become an ion.
The understanding that the movement of electrons, rather than protons, creates
electrical effects is a relatively new discovery. Unfortunately, this means that many
of the rules and conventions used when talking about electricity refer to electric
current as the movement of positive charge carriers. Electricity and circuits will be
discussed in greater detail in Chapter 13.
MEASURING CHARGE
In order to measure the actual amount of charge on a charged object, a ‘natural’
unit would be the magnitude (size) of the charge on one electron or proton. This
fundamental charge is often referred to as the elementary charge and is given the
symbol q. A proton therefore has a charge of qp and an electron has a charge of qe.
The size of the elementary charge is very small. For most practical situations, it
is more convenient to use a larger unit to measure charge. The SI (standard) unit of
charge is known as the coulomb (symbol C). It is named after Charles-Augustin
de Coulomb, who was the first scientist to measure the forces of attraction and
repulsion between charges.
A coulomb is quite a large unit of charge: +1 coulomb (1 C) is equivalent to the
combined charge of 6.2 × 1018 protons. Therefore, the charge on a single proton is
+1.6 × 10−19 C. Similarly, −1 C is equivalent to the combined charge of 6.2 × 1018
electrons and the charge on a single electron is −1.6 × 10−19 C. The letter q is used to
represent the amount of charge.
Worked example 12.1.1
THE AMOUNT OF CHARGE ON A GROUP OF ELECTRONS
Charge Positive Negative
Positive repel
attract
Negative attract
repel
An excess of electrons causes an
object to be negatively charged,
and a deficit in electrons will mean
the object is positively charged.
PHYSICSFILE ICT
Electron models
The way an electron moves around the
nucleus of an atom is more complex
than the simple planetary model would
suggest. An individual electron is so
small that its exact position at any
point in time is impossible to measure.
Recent models of the structure of the
atom describe an electron in terms
of the probability of finding it in a
certain location. In diagrams of atoms,
this is often represented by drawing
the electrons around the nucleus as
a fuzzy cloud, rather than points or
solid spheres.
nucleus
electron cloud
proton
neutron
FIGURE 12.1.2 The nucleus of an atom
occupies about 10−12 of the volume of the
atom, yet it contains more than 99% of its
mass. Atoms are mostly empty space.
Calculate the charge, in coulombs, carried by 6 billion electrons.
Thinking
Express 6 billion in scientific notation.
Working
1 billion = 1 × 109
6 billion = 6 × 109
Calculate the charge, q, in coulombs by
q = (6 × 109) × (qe)
multiplying the number of electrons by
= (6 × 109) × (−1.6 × 10−19 C)
the charge on an electron (−1.6 × 10−19 C).
= −9.6 × 10−10 C.
The elementary charge, qp, of a
proton is equal to 1.6 × 10−19 C.
The elementary charge, qe,
of an electron is equal to
−1.6 × 10−19 C.
Worked example: Try yourself 12.1.1
THE AMOUNT OF CHARGE ON A GROUP OF ELECTRONS
Calculate the charge, in coulombs, carried by 4 million electrons.
CHAPTER 12 | ELECTROSTATICS
339
Worked example 12.1.2
THE NUMBER OF ELECTRONS IN A GIVEN AMOUNT OF CHARGE
The net charge on an object is −3.0 μC (1 μC = 1 microcoulomb = 1 × 10−6 C).
Calculate the number of extra electrons on the object.
Thinking
Working
Express −3.0 μC in scientific notation.
q = −3.0 μC
= −3.0 × 10−6 C.
Find the number of electrons by
dividing the charge on the object by the
charge on an electron (−1.6 × 10−19 C).
ne =
q
qe
=
−3.0 × 10−6
−1.6 × 10−19
= 1.9 × 1013 electrons.
Worked example: Try yourself 12.1.2
THE NUMBER OF ELECTRONS IN A GIVEN AMOUNT OF CHARGE
The net charge on an object is −4.8 μC (1 μC = 1 microcoulomb = 1 × 10−6 C).
Calculate the number of extra electrons on the object.
ELECTRICAL CONDUCTORS AND INSULATORS
Electrons are much easier to move than protons. They also move more freely in
some materials than in others.
In some materials, the electrons are only very slightly attracted to their respective
nuclei. These materials are known as metals. Metals are conductors of electricity.
In conductors, loosely held electrons can effectively ‘jump’ from one atom to another
and move freely throughout the material. This can be seen in Figure 12.1.3.
flow of electrons
nucleus
electron
wire
FIGURE 12.1.3 Electrons moving through a conductor. The electrons are free to move throughout the
lattice of positive ions.
FIGURE 12.1.4 These copper wires conduct
electricity by allowing the movement of charged
particles.
340
Copper is an example of a very good conductor. For this reason it is used in
telecommunications and electrical and electronic products (Figure 12.1.4).
By comparison, the electrons in non-metals are very tightly bound to their
respective nuclei and cannot readily move from one atom to another. Non-metals
do not conduct electricity very well and are known as insulators. Some common
conductors and insulators are listed in Table 12.1.1.
MODULE 4 | ELECTRICITY AND MAGNETISM
PHYSICS IN ACTION
ICT
Lightning
T = −65°C
Lightning is one of nature’s greatest
spectacles. No wonder it was so long thought
of as the voice of the gods. In the mideighteenth century, Benjamin Franklin showed
that lightning is basically the same sort of
electrical phenomenon as can be achieved by 12 km
rubbing a glass rod with wool, or rubbing a
balloon on your hair.
A typical lightning bolt transfers 10 or
more coulombs of negative charge (over
60 billion billion electrons) in approximately
one thousandth of a second. A moderate
thundercloud with a few flashes per minute
generates several hundred megawatts of
electrical power, the equivalent of a small
power station.
T = −15°C
5 km
T = 10°C
It is thought that, during a thunderstorm,
charge is transferred in collisions between the
2 km
tiny ice crystals that form as a result of the
cooling of moist air flowing upwards and the
larger, falling hailstones. As a result of small
temperature differences between the crystals
FIGURE 12.1.5 A thundercloud can be several kilometres wide and well over 10 km high.
and hailstones, the crystals become positively
charged and the hailstones negatively charged. Strong updrafts drive the electrical processes that lead to the separation of charge. The
strong negative charge of the lower region of the cloud induces positive charges on tall
The crystals carry their positive charge to the
objects on the ground. This may lead to a discharge, which can form a conductive path
top of the cloud while the negative charge
for lightning.
accumulates in the lower region. There is
normally also a second smaller positively charged region at
Because of the electric field, the free electrons and ions
the bottom owing to positive charges attracted up from the
gain kinetic energy and collide with more molecules,
ground towards the negative region, as seen in Figure 12.1.5.
starting an ‘avalanche of charges’. This is the lightning
There are strong electric fields between the regions of
opposite charge. If they become strong enough, electrons
are stripped from the air molecules and become ionised.
TABLE 12.1.1
flash seen either within the cloud or between the Earth
and the cloud. Most flashes are within the cloud; only a
relatively small number actually strike the ground.
Some common conductors and insulators.
Conductors
Insulators
Good
all metals, especially silver, gold, copper
and aluminium
any ionic solution
plastics
polystyrene
dry air
glass
porcelain
cloth (dry)
Moderate
tap water
earth
semiconductors, e.g. silicon, germanium
wood
paper
damp air
ice, snow
CHAPTER 12 | ELECTROSTATICS
341
ELECTROSTATIC FORCES
All electrically charged objects are surrounded by an electric field. In an electric
field, a charged object experiences a force on it. Electrically charged particles also
produce a force on other objects. In both cases, this force is called the electrostatic
force.
The electrostatic force, like all forces, is expressed in newtons. Electrostatic force
is a vector that has both direction and magnitude.
The direction of the electrostatic force between two objects is determined by
the rules of charge interaction:
• Objects with the same charge repel each other. The electrostatic force is
repulsive, and the objects will be pushed away from each other.
• Objects with a different charge attract each other. The electrostatic force is
attractive, and the objects will be pulled towards each other.
PA
4.1
PA
4.1
WS
4.1
WS
4.2
The magnitude of the electrostatic force can be determined using Coulomb’s
law, which is discussed in greater detail in Section 12.3. Electrostatic force is an
example of a non-contact force (or a force mediated by a field).
+ ADDITIONAL
Semiconductors
Some materials, such as silicon, are known as semimetals
or metalloids. Their properties are somewhere between
those of metals and non-metals. For example, the electrons
in a silicon atom are not as tightly bound to the nucleus as
those of a non-metal but they are not as easy to remove as
the electrons in a metal. Hence, silicon and elements like it
are known as semiconductors.
The ability of silicon to conduct electricity can be adjusted
by adding small amounts of other elements such as
boron, phosphorus, gallium or arsenic in a process known
as doping. Adding another substance contributes free
electrons, which greatly increases the conductivity of silicon
in electronic devices. This makes silicon very useful in the
construction of computer chips (or integrated circuits).
Current manufacturing technologies allow over a
billion transistors and other electronic components
to be fabricated on an integrated circuit. Much of the
convenience of our modern lifestyle is based on the unique
conductive properties of silicon, and it has many electronic
applications. Some of them are shown in Figure 12.1.6.
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MODULE 4 | ELECTRICITY AND MAGNETISM
FIGURE 12.1.6
construction.
All of these electronic devices use silicon in their
12.1 Review
SUMMARY
• The letter q is used to represent the amount of
charge. The SI unit of charge is the coulomb (C).
• Electrons move easily through conductors, but
not through insulators. This is because the
electrons in materials that are good conductors
are weakly attracted to the nucleus, and electrons
in insulators are more strongly attracted to the
nucleus.
• The charge of a proton (qp) is equal to
1.6 × 10−19 C. The charge of an electron (qe) is
−1.6 × 10−19 C.
• Charged objects produce an electrostatic force
on other objects that causes like charges to repel,
and opposite charges to attract.
• Like charges repel; unlike charges attract.
• When an object loses electrons, it develops a
positive net charge; when it gains electrons, it
develops a negative net charge.
KEY QUESTIONS
1
Plastic strip A, when rubbed, is found to attract plastic
strip B. Strip C is found to repel strip B. What will
happen when strip A and strip C are brought close
together?
2
Which of the following statements is correct?
A The electrostatic force between two positive charges
is attractive.
B The electrostatic force between two negative
charges is attractive.
C The electrostatic force between a positive and
negative charge is attractive.
D The electrostatic force between a positive and
negative charge is repulsive.
3
If an atom has gained an electron, it becomes:
A a molecule
B an ion
C a particle
D a neutron.
4
Calculate how many electrons make up a charge of
−5.0 C.
5
Calculate the charge, in coulombs, of 4.2 × 1019
protons.
6
Explain why electric circuits often consist of wires that
are made from copper and are coated in protective
rubber.
CHAPTER 12 | ELECTROSTATICS
343
12.2 Electric fields
GO TO ➤ Section 4.1, page 120
A field is a region of space where objects experience a force due to a physical
property related to the field. An example is the earth’s gravitational field. Objects fall
to the ground as they are affected by the force due to earth’s gravity. In this section,
the electric field will be explained.
An electric field surrounds positive and negative charges, and exerts a force
on other charges within the field. The electric field around charged objects can be
represented by field lines, as shown in Figure 12.2.1.
ELECTRIC FIELDS
FIGURE 12.2.1 Charged plasma follows lines of
the electric field produced by a Van de Graaff
generator.
There are four fundamental forces in nature that act at a distance, called noncontact forces, as described in Section 4.1. That is, they can exert a force on an
object without making any physical contact with it. These include the strong nuclear
force, the weak nuclear force, the electromagnetic force and the gravitational force.
In order to understand these forces, scientists use the idea of a field. A field is
a region of space around an object that has certain physical properties, such as
electric charge or mass. Another object with that physical property in the field will
experience a force without any contact between the two objects.
For example, there is a gravitational field around the Earth due to the mass
of the Earth. Any object with mass that is located within this gravitational field
experiences a force of attraction towards the Earth. An example of this is shown
in Figure 12.2.2.
Similarly, any charged object has a region of space around it (an electric field)
where another charged object will experience a force. This is one aspect of the
electromagnetic force. Unlike gravity, which only exerts an attractive force, electric
fields can exert forces of attraction or repulsion.
ELECTRIC FIELD LINES
The gravitational field of the
Earth applies a force on the skydiver.
FIGURE 12.2.2
An electric field is a vector quantity, which means it has both direction and strength.
In order to visualise electric fields around charged objects you can use electric
field lines. Some field lines are already visible—for example the girl’s hair in
Figure 12.2.3 is tracing out the path of the field lines. Diagrams of field lines can
also be constructed.
Field lines are drawn with arrowheads on them indicating the direction of the
force that a small positive test charge would experience if it were placed in the
electric field. Therefore, field lines point away from positively charged objects
and towards negatively charged objects. Usually, only a few representative lines
are drawn.
Remember: like charges repel and unlike (opposite) charges attract.
The girl’s hair follows the lines
of the electric field produced when she became
charged while sliding down a plastic slide.
FIGURE 12.2.3
The density of field lines (how close they are together) is an indication of the
relative strength of the electric field. This is explained in more detail later in this
section.
Rules for drawing electric field lines
When drawing electric field lines (in two dimensions) around a charged object there
are a few rules that need to be followed.
• Electric field lines go from positively charged objects towards negatively charged
objects.
• Electric field lines start and end at 90° to the surface, with no gap between the
lines and the surface.
• Field lines can never cross; if they did it would indicate that the field is in two
directions at that point, which can never happen.
344
MODULE 4 | ELECTRICITY AND MAGNETISM
•
•
•
•
•
•
Around small charged spheres, called point charges, the field lines radiate like
spokes on a wheel.
Around point charges you should draw at least eight field lines: top, bottom, left,
right and another field line in between each of these.
Between two point charges, the direction of the field at any point is the resultant
field vector determined by adding the field vectors due to each of the two
point charges.
Between two equal and oppositely charged point charges called a (dipole), the
field lines are drawn from the positive charge to the negative charge.
Between two oppositely charged parallel plates, the field lines are evenly spaced
and are drawn straight from the positive plate to the negative plate.
Always remember that these drawings are two-dimensional representations of a
three-dimensional field.
Figure 12.2.4 shows some examples of how to draw electric field lines.
+
A single positive charge
Two positive
charges
A positive and a
negative charge
(dipole)
+
+
+
–
+ + + + + + + + + + + + +
Parallel oppositely
charged plates
– – – – – – – – – – – – –
FIGURE 12.2.4 Grass seeds suspended in oil align themselves with the electric field. The diagram next
to each photo shows lines representing the electric field.
Strength of the electric field
The distance between adjacent field lines indicates the strength of the field. Around
a point charge, the field lines are closer together near the charge and get further
apart as you move further away. You can see this in the field-line diagrams in
Figure 12.2.4. Therefore, the electric field strength decreases as the distance
from a point charge increases.
WS
4.3
CHAPTER 12 | ELECTROSTATICS
345
A uniform electric field is established between two parallel metal plates that are
oppositely charged. The field strength is constant at all points within a uniform
electric field, so the field lines are parallel.
FORCES ON FREE CHARGES IN ELECTRIC FIELDS
+
+
+
+
+
+
F⃗
⊝
⊕
F⃗
−
+
−
−
−
−
E⃗
−
−
FIGURE
12.2.5 The direction of the electric
field (E) indicates the direction in which a force
would act on a positive charge. A negative
charge would experience a force in the opposite
direction to the field.
If a charged particle, such as an electron, were placed within an electric field, it
would experience a force. The direction of the field and the sign of the charge allow
you to determine the direction of the force.
Figure 12.2.5 shows a positive test charge (proton) and a negative test charge
(electron) within a uniform electric field. The direction of an electric field is
defined as the direction of the force that a positive charge would experience within
the electric field. So, an electron will experience a force in the opposite direction
to the electric field, while a proton will experience a force in the same direction
as the field.
The force experienced by a charged particle due to an electric field can be
determined using the equation:
F = qE
where:
F is the force on the charged particle (N)
q is the charge of the object experiencing the force (C)
E is the strength of the electric field (N C−1).
This equation illustrates that the force
experienced by a charge is proportional
to the strength of the electric field, E , and the size of the charge, q. The force on
the charged particle causes it to accelerate in the field. This means that the particle
could increase its velocity, decrease its velocity or change its direction when it is in
the electric field.
To calculate the acceleration of the particle due to this force acting on it, you can
use the equation from Newton’s second law:
F = ma
where:
m is the mass of the accelerating particle (kg)
a is the acceleration (m s−2).
SKILLBUILDER
N
Magnitude
As you know from Chapter 8, a
positive or a negative sign outside
the vector indicates its direction.
Sometimes, though, you want
to know only the magnitude of
a particular variable but not its
direction. In that case, the negative
sign may be ignored.
Generally, when you are looking
for the magnitude the variable will
be written as a scalar quantity, i.e.
F instead of F. You may also see
this written as the absolute
value
of a variable, such as | F |.
346
PHYSICSFILE ICT
Bees
Bees are thought to
use electric fields to
communicate, find food
and avoid flowers that have
been visited by another bee
recently. Their antennae are
bent (deflected) by electric
fields and they can sense
the amount of deflection
(Figure 12.2.6).
The charge that builds up
FIGURE 12.2.6 A bee can detect changes in the electric field
on their bodies helps them
around its body.
collect pollen grains and
transport them to other
flowers. The altered electric field around a flower that has recently been visited is a
signal to other bees to find food elsewhere.
MODULE 4 | ELECTRICITY AND MAGNETISM
Worked example 12.2.1
CALCULATING ELECTRIC FIELD STRENGTH
Calculate the uniform electric field that would cause a force of 5.00 × 10−21 N on
an electron (qe = −1.602 × 10−19 C).
Thinking
Rearrange the relevant equation
to make electric field strength the
subject.
Substitute the values for F and q into
the rearranged equation and calculate
the answer.
Working
F = qE
E = Fq
E=
=
F
q
5.00 × 10−21
−1.602 × 10−19
−2
=3
.12 × 10 N C−1 in the opposite
direction to the force
Worked example: Try yourself 12.2.1
CALCULATING ELECTRIC FIELD STRENGTH
Calculate the uniform electric field that creates a force of 9.00 × 10−23 N on a
proton (qp = +1.602 × 10−19 C).
WORK DONE IN UNIFORM ELECTRIC FIELDS
Electrical potential energy is a form of energy that is stored in a field. The concept of
work was discussed in Chapter 5. Work is done on the field when a charged particle
is forced to move in the electric field. Conversely, when energy is stored in the
electric field, then work can be done by the field on the charged particle.
Electrical potential (V ) is defined as the work required per unit charge
to move a positive point charge from infinity to a place in the electric field. The
electrical potential at infinity is defined as zero. Remember that the work done by an
object is equal to the change in energy, so the electrical potential energy can also be
defined as the difference in electric potential energy (ΔU ) per unit charge.
This definition leads to the equation:
V = ∆qU = Wq
where:
U is the potential energy (in J)
W is the work done on a positive point charge or on the field (in J)
q is the charge of the point charge (in C)
V is the electrical potential (in J C−1) or volts (V),
Consider two parallel plates, as shown in Figure 12.2.7, in which the positive
plate is at a potential (V ) and the other plate is earthed, which is defined as zero
potential. The difference in potential between these two plates is called the potential
difference (V ).
Between any two points in an electric field ( E ) separated by a distance (d ) that
is parallel to the field, the potential difference V is then defined as the change in the
electrical potential between these two points. See Figure 12.2.8.
GO TO ➤ Section 5.2, page 155
V +
+
+
+
+
+
+
E⃗
V=0
FIGURE 12.2.7 The potential of two plates when
one has a positive potential and the other is
earthed.
V +
+
+
d1
+
V1
+
d2
+
V2
+
E⃗
V=0
FIGURE 12.2.8 The potential difference between
two points in a uniform electric field.
CHAPTER 12 | ELECTROSTATICS
347
Electric field strength
From Chapter 5, recall that work is equal to the product of the force and the
displacement:
W = F s =Fd
The electrostatic force is expressed as:
F = qE
and as explained above, the potential V is equal to:
V =
∆U
q
= Wq
Putting these three equations together:
W = qV = Fd
qV = qEd
∴V = Ed
This is often rearranged as the equation:
E = − V
d
where:
V is the electrical potential (V)
E is the electrical field strength (V m−1)
d is the displacement or distance between points, parallel to the field (m).
The negative value is used because the electric field E is always from the higher
potential towards the lower potential, but in this context the change in potential
energy becomes negative.
CALCULATING WORK DONE
By combining the magnitudes of the two equations mentioned so far, you can derive
an equation for calculating the work done on a point charge to move it through a
distance d across a potential difference:
W = qEd
where:
W is the work done on the point charge, or on the field (J)
q is the charge of the point charge (C)
E is the magnitude of the electrical field strength (V m−1 or N C−1)
d is the distance between the points, parallel to the field (m).
Work done by or on an electric field
When calculating work done, which changes the electrical potential energy,
remember that work is done either:
• by the electric field on a charged object, or
• on the electric field by forcing the object to move.
You need to examine what is happening in a particular situation to know how
the work is being done.
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MODULE 4 | ELECTRICITY AND MAGNETISM
For example, if a charged object is moving in the direction it would naturally
tend to go within an electric field, then work is done by the field. So when a positive
point charge is moved in the direction of the electric field, the electric field has done
work on the point charge. (Refer to q2 in Figure 12.2.9.)
When work is done by a charged object on an electric field, the object is forced
to move against the direction it would naturally go. Work has been done on the
field by forcing the object to move against the field. For example, if a force causes
a positive charge to move towards the positive plate within a uniform electric field,
work has been done on the electric field by forcing the object to move. (See q1 in
Figure 12.2.9.)
If a charge doesn’t move any distance parallel to the field, then no work is done
on or by the field. If a charge moves perpindicularly to the field, then no work is
done. (See q3 in Figure 12.2.9.)
V +
+
+
d1
⊕ q1
V=0
+ +
⊕ q2
d2
⊕
q3
+
d3
+
E⃗
FIGURE 12.2.9 Work is being done on the field
by moving q1 and work is being done by the
field on q2. No work is done on q3, since it is
moving perpendicular to the field.
Worked example 12.2.2
WORK DONE ON A CHARGE IN A UNIFORM ELECTRIC FIELD
A student sets up a parallel plate arrangement so that one plate is at a
potential of 12.0 V and the other earthed plate is positioned 0.50 m away
(qp = +1.602 × 10−19 C).
Calculate the work done to move a proton from the earthed plate a distance of
10.0 cm towards the negative plate. In your answer, identify what does the work
and what the work is done on.
Thinking
Working
Identify the variables presented in the
problem to calculate the electric field
strength.
V1 = 0 V
Use the equation E = − V to determine
E = − V
d
V2 = 12.0 V
d = 0.50 m
d
the electric field strength.
= − 0 − 12
0.50
= 24.0 V m−1
Use the equation W = qEd to determine
the work done. Note that d here is the
distance that the proton moves.
W = qEd
Determine if work is done on the
charge by the field or if work is done
on the field.
As the positively charged proton is
moving naturally towards the negative
plate then work is done on the proton
by the field.
= 1.602 × 10−19 × 24.0 × 0.100
= 3.84 × 10−19 J
Worked example: Try yourself 12.2.2
WORK DONE ON A CHARGE IN A UNIFORM ELECTRIC FIELD
A student sets up a parallel plate arrangement so that one plate is at a potential
of 36.0 V and the other earthed plate is 2.00 m away.
Calculate the work done to move an electron a distance of 75.0 cm towards the
negative plate (qe = −1.602 × 10−19 C).
In your answer, identify what does the work and what the work is done on.
PA
4.2
CHAPTER 12 | ELECTROSTATICS
349
SKILLBUILDER
N
Using units in an equation to check
for dimensional consistency
Scientists know that each term in an equation stands for a particular quantity.
The position of the term in the equation tells you where that quantity should
go. The units used to measure that quantity are not used in the calculations.
Units are only indicated on the final line of the solved equation.
For example, the equation for the area (A) of a rectangle of length (L) and
width (W) is A = L × W. For a rectangle 7 m long and 4 m wide, the calculation
is written:
A = L × W = 7 × 4 = 28 m2
You can use units to check the dimensional consistency of the answer. In
the example above, the two quantities L (length) and W (width) both have to
be expressed in consistent units, in this case metres (m), to give an answer
that is expressed in square metres (m × m = m2).
If you had made a mistake, and used the formula A = L + W instead, the
answer would be expressed in metres only. This is not the correct unit for
area, so you would know that was wrong.
Similarly, you can equate different units using dimensional analysis. For
example,
if you combine the equations for the magnitude of the electric field
V
|E| = d and E = Fq , you end up with Vd = Fq . Looking at the units for each side of
this equation, V m−1 must equal N C−1.
PHYSICS IN ACTION
ICT
Potential difference
If repairs are required to high voltage transmission lines,
it might not be possible to disrupt electricity supply to the
surrounding neighbourhood. This is particularly true for
repairs to high voltage transmission lines servicing major
metropolitan areas. To avoid disrupting the electricity
supply, repairs can be done on live wires.
Just tens of milliamps of electric current can cause
electrocution. Electric current flows through a circuit
when there is a potential difference across the circuit. If
two points are at the same electric potential, there is no
potential difference, and as a result no current will flow.
One technique for repairing live high-voltage power lines
is for the worker to come in contact with the wire. This puts
the worker at the same electric potential as the wire. This
may be done by touching the wire directly (Figure 12.2.10),
or by using an elevated platform or ladder that comes in
contact with the wire. There is then no potential difference
between the worker and the wire, so no current will flow
and the repairs can be carried out safely.
350
MODULE 4 | ELECTRICITY AND MAGNETISM
FIGURE 12.2.10 Workers repairing a live high voltage power line. There
is no potential difference between the workers and the wires, therefore
no current will flow and the repairs can be safely carried out.
12.2 Review
SUMMARY
• An electric field is a region of space around a
charged object in which another charged object
will experience a force.
• When charges are in an electric field, they accelerate
in the direction of the force acting on them.
• Electric fields are represented using field lines.
• The force on a charged particle can be
determined using the equation F = qE.
• Electric field lines point in the direction of the force that
a positive charge within the field would experience.
• The force is related to the acceleration of a particle
using the equation F = ma.
• A positive charge experiences a force in the direction
of the electric field and a negative charge experiences
a force in the opposite direction to the field.
• Electrical potential energy is stored in an electric field.
• Electrical potential (V ) can be determined using
the equation V = ∆qU . This is equivalent to the work
done per unit charge.
• The spacing between the field lines indicates the
strength of the field. The closer together the lines
are, the stronger the field.
• Electric field strength can be expressed as E = Fq
and also as E = − V .
• When a charged object is moved against the
direction it would naturally move in an electric
field, then work is done on the field.
• When a charged object is moved in the direction
it would naturally tend to move in an electric field,
then the field does work on the particle.
d
• Around point charges the electric field radiates in
all directions (three-dimensionally).
• The work done on or by an electric field can be
calculated using the equations W = qV or W = qEd.
• Between two oppositely charged parallel plates,
the field lines are parallel and therefore the field
has a uniform strength.
KEY QUESTIONS
1
2
3
Which of the following options correctly describes an
electric field?
A a region around a charged object that causes a
charge on other objects in that region
B a region around a charged object that causes a
force on other objects in that region
C a region around a charged object that causes a
force on other charged objects in that region
D a region around an object that causes a force on
other objects in that region.
Which of the following options correctly defines the
direction of an electric field?
A away from a negatively charged object
B away from a positively charged object
C away from a neutrally charged object
D towards a positively charged object
Identify whether the rules below for drawing electric
field lines are true or false:
a Electric field lines start and end at 90° to the surface,
with no gap between the lines and the surface.
b Field lines can cross; this indicates that the field is
in two directions at that point.
c Electric fields go from negatively charged objects
to positively charged objects.
d Around small charged spheres called point charges
you should draw at least eight field lines: top,
bottom, left, right and between each of these.
4
5
6
7
8
e Around point charges the field lines radiate like
spokes on a wheel.
f Between two point charges the direction of the field
at any point in the field is due to the closest of the
two point charges.
g Between two oppositely charged parallel plates the
field between the plates is evenly spaced and is drawn
straight from the negative plate to the positive plate.
Calculate the magnitude of the force applied to a
balloon carrying a charge of 5.00 mC in a uniform
electric field of 2.50 N C−1.
Calculate the charge on a plastic disc if it experiences
a force of 0.025 N in a uniform electric field of 18 N C−1.
Calculate the magnitude of the acceleration of an
electron in a uniform electric field of 3.25 N C−1. (The
mass of an electron is 9.11 × 10−31 kg and its charge is
−1.602 × 10−19 C.)
Calculate the potential difference that exists between
two points separated by 30.0 cm, parallel to the field
lines, in an electric field of strength 4000 V m−1.
Using
dimensional
analysis, show that the units for
E = − V and E = Fq equate. The following relationships
d
may be helpful:
V = J C−1
N = kg m s−2
J = kg m2 s−2.
CHAPTER 12 | ELECTROSTATICS
351
12.3 Coulomb’s law
Electricity is one of nature’s fundamental forces. It was Charles Coulomb, in 1785,
who first published the quantitative details of the force that acts between two
electric charges. The force between any combination of electrical charges can be
understood in terms of the force between two ‘point charges’ separated by a certain
distance, as seen in Figure 12.3.1. The effect of distance on the electric field strength
from a single charge and the force created by that field between charges is explored
in this section.
F⃗
F⃗
Two similarly charged balloons will repel each other by applying a force on each other.
FIGURE 12.3.1
THE FORCE BETWEEN CHARGED PARTICLES
GO TO ➤ Skillbuilder, page 254
Coulomb found that the force between two point charges (q1 and q2) separated
by a distance (r) was proportional to the product of the two charges, and inversely
proportional to the square of the distance between them.
This is another example of an inverse square law, as discussed in Chapter 9.
Coulomb’s law can be expressed by the following equation:
1 q1q2
F = 4 πε
2
0
r
where:
F is the force on each charged object (in N) (the electrostatic force)
q1 is the charge on one point (in C)
q2 is the charge on the other point (in C)
r is the distance between each charged point (in m)
ε0 is the permittivity of free space, which is equal to 8.8542 × 10−12 C2 N−1 m−2
in air or in a vacuum.
The sign of the charges in the calculation indicates the direction of the force.
A positive force value indicates repulsion and a negative force value indicates
attraction.
The above value of the permittivity of free space (ε0) is constant for air or
1
a vacuum, so the expression 4 πε
in Coulomb’s law equation can be calculated.
0
The result of the calculation is given the name Coulomb’s constant (k) and is
equal to 8.9875 × 109 N m2 C−2. For ease of calculation, this is usually rounded
to two significant figures, as 9.0 × 109 N m2 C−2. So, in Coulomb’s law, the
equation becomes:
qq
F = k 12 2
r
where k = 9.0 × 109 N m2 C−2.
352
MODULE 4 | ELECTRICITY AND MAGNETISM
PHYSICSFILE ICT
Static electricity
Static electricity is a phenomenon you may
have experienced when you felt an electric
shock, such as when touching an object.
The static shock occurs when one material
or object that is negatively charged comes
in contact with another material or object
that is positively charged.
You may have experienced a static shock
when walking across a carpet, then
touching a door handle. Static electricity
is generated when the soles of your shoes
rub against the carpet. When you touch the
door handle, the electrons are transferred
to the door handle, which causes you to
feel a shock.
Static electric shocks usually result in a
quick jolt that startles you, with no harmful
effects. There are cases, though, where
static electricity and static electric shocks
can be dangerous or cause damage. For
instance, with electronic circuits, a static
shock may be powerful enough to destroy
electronic components. Anyone working on
electronic components (e.g. manufacturing,
repairing or testing) must ensure they are
using antistatic devices. A common device
is a wrist strap, which ensures the person’s
body is electrically grounded and prevents
any build-up of static charge (Figure 12.3.2).
FIGURE 12.3.2 When building, testing or
repairing electronic circuits, technicians use a
wrist strap to ground their body, to ensure no
static electricity is transferred to the electronic
circuit.
FACTORS AFFECTING THE ELECTRIC FORCE
The force between two charged points is proportional to the product of the two
charges, as seen in Figure 12.3.3. If there is a force of 10 N between two charged
points and either charge was doubled, then the force between the two points would
increase to 20 N. It is interesting to note that, regardless of the charge on each point,
the forces on each point in a pair are the same. For example, if q1 is +10 μC and q2 is
+10 μC, then the repulsive force on each of these points is equal in magnitude. The
force is also equal on each point if q1 is +100 μC and q2 is +1 μC.
The force is also inversely proportional to the square of the distance between the
two charged points. This means that if the distance between q1 and q2 is doubled, the
force on each point charge is reduced to one-quarter of its previous value.
ONE COULOMB IN PERSPECTIVE
F⃗
q1
F⃗
r
q2
F⃗
F⃗
q1
FIGURE 12.3.3
charges.
r
q2
Forces acting between two point
Using Coulomb’s law you can calculate the force between two charges of 1 C each,
placed 1 m apart. The force is 9.0 × 109 N, or approximately 1 × 1010 N. This is
equivalent to the weight provided by a mass of 918 000 tonnes (Figure 12.3.4).
This demonstrates that a 1 C charge is a huge amount of charge. In reality,
the amount of charge that can be placed on ordinary objects is a tiny fraction
of a coulomb. Even a highly charged Van de Graaff generator has only a few
microcoulombs (1 μC = 1 × 10−6 C) of excess charge.
10
FIGURE 12.3.4 Two 1 C charges 1 m apart would produce a force of 1 × 10 N, which is almost twice
the weight of the Sydney Harbour Bridge.
CHAPTER 12 | ELECTROSTATICS
353
Worked example 12.3.1
USING COULOMB’S LAW TO CALCULATE FORCE
Two small spheres A and B act as point charges separated by 10.0 cm in air.
Calculate the force on each point charge if A has a charge of 3.00 μC and B has
a charge of −45.0 nC. (Use ε0 = 8.8542 × 10−12 C2 N−1 m−2.)
Thinking
Working
Convert all values to SI
units.
qA = 3.00 × 10−6 C
qB = −45.0 × 10−9 = −4.50 × 10−8 C
r = 0.100 m
1 q1q2
F = 4 πε
2
State Coulomb’s law.
0
Substitute the values for
qA, qB, r and ε0 into the
equation and calculate
the answer.
Assign a direction based
on a negative force being
attraction and a positive
force being repulsion.
F=
r
1
4 π × 8.8542 × 10−12
× 3.00 × 10
−6
× −4.50 × 10−8
0.1002
= −0.121 N
F = 0.121 N attraction
Worked example: Try yourself 12.3.1
USING COULOMB’S LAW TO CALCULATE FORCE
Two small spheres A and B act as point charges separated by 75.0 mm in air.
Calculate the force on each point charge if A has a charge of 475 nC and B has a
charge of 833 pC. (Use ε0 = 8.8542 × 10−12 C2 N−1 m−2.)
Worked example 12.3.2
USING COULOMB’S LAW TO CALCULATE CHARGE
Two small positive point charges with equal charge are separated by 1.25 cm in air.
Calculate the charge on each point charge if there is a repulsive force of 6.48 mN
between them. (Use k = 9.0 × 109 N m2 C−2.)
Thinking
Convert all values to SI units.
Working
F = 6.48 × 10−3 N
State Coulomb’s law.
r = 1.25 × 10−2 m
qq
F = k 122
Substitute the values for F, r and
k into the equation and calculate
the answer.
r
q1q2 =
=
Fr 2
k
6.48 × 10−3 × (1.25 × 10−2 )2
9.0 × 109
= 1.125 × 10−16
Since q1 = q2:
q12 = 1.125 × 10−16
∴ q1 = 1.125 × 10−16
= +1.06 × 10−8 C
q2 = +1.06 × 10−8 C
354
MODULE 4 | ELECTRICITY AND MAGNETISM
Worked example: Try yourself 12.3.2
USING COULOMB’S LAW TO CALCULATE CHARGE
Two small point charges are charged by transferring a number of electrons from
q1 to q2, and are separated by 12.7 mm in air. The charges at the two points are
equal and opposite.
Calculate the charge on q1 and q2 if there is an attractive force of 22.5 μN
between them.
(Use k = 9.0 × 109 N m2 C−2.)
THE ELECTRIC FIELD AT A DISTANCE FROM A CHARGE
In Section 12.2, the electric field, E, was defined as being proportional to the force
exerted on a positive test charge and inversely proportional to the magnitude of that
charge. The electric field is measured in N C−1.
It is useful also to be able to determine the electric field at a distance from a
single point charge:
The magnitude of the electric field at a distance from a single point charge is
given by:
E = k q2
r
where:
E is the magnitude of the of the electric field strength around a point (in N C−1)
q is the charge on the point creating the field (in C)
r is the distance from the charge (in m)
k = 9.0 × 109 N m2 C−2.
The magnitude of E that is determined is independent of the value of the test
charge and depends only on the charge, q, producing the field. This formula can
also be referred to as Coulomb’s law, in this case for determining the magnitude of
the electric field produced by a point charge.
Worked example 12.3.3
ELECTRIC FIELD OF A SINGLE POINT CHARGE
Calculate the magnitude and direction of the electric field at a point P at a
distance of 20 cm below a negative point charge, q, of 2.0 × 10−6 C.
Thinking
Working
Convert units to SI units as required.
q = −2.0 × 10−6 C
r = 20 cm = 0.20 m
Substitute the known values to find the
magnitude of E using:
E =k
E =k
= 9.0 × 109 × 2.0 × 102
q
r2
The direction of the field is defined as
that acting on a positive test charge
(see previous section). Point P is below
the charge.
q
r2
−6
0.20
−1
= 4.5 × 105 N C
Since the charge is negative, the
direction will be toward the charge q,
or upwards.
Worked example: Try yourself 12.3.3
ELECTRIC FIELD OF A SINGLE POINT CHARGE
Calculate the magnitude and direction of the electric field at point P at a
distance of 15 cm to the right of a positive point charge, q, of 2.0 × 10−6 C.
WS
4.4
CHAPTER 12 | ELECTROSTATICS
355
12.3 Review
SUMMARY
• Coulomb’s law for the force between two charges
q1 and q2 separated by a distance r is:
1 q1q2
F = 4 πε
2
0
• The magnitude of the electric field, E, at a distance
r from a single point charge q is given by:
E =k
r
q
,
r2
where k = 9.0 × 109 N m2 C−2
• The constant, ε0, in Coulomb’s law has a value of
8.8542 × 10−12 C2 N−1 m−2.
• For air or a vacuum, the expression
1
4 πε 0
at the
front of Coulomb’s law can be simplified to the
value of k, called Coulomb’s constant, which has a
value of approximately 9.0 × 109 N m2 C−2. So:
F = k q1rq22
KEY QUESTIONS
1
Choose the responses that correctly complete the
following table.
Force
q1 charge q2 charge
A
positive
positive
B
negative positive/
negative
positive
C
positive
positive/negative attraction/repulsion
D
negative positive/
negative
2
3
356
negative
The electric field is being measured at point P at a
distance of 5.0 cm from a positive point charge, q, of
3.0 × 10−6 C. What is the magnitude of the field at P to
two significant figures? (Use k = 9.0 × 109 N m2 C−2.)
5
Calculate the magnitude of the force that would exist
between two point charges of 1.00 C each, separated
by 1.00 km. (Use k = 9.0 × 109 N m2 C−2.)
6
A point charge of 6.50 mC is suspended from a
ceiling by an insulated rod. At what distance from the
point charge will a small sphere of mass 10.0 g with
a charge of −3.45 nC be located if it is suspended in
air? Use k = 9.0 × 109 N m2 C−2, and remember that the
magnitude of the weight force is equal to F = mg.
7
A charge of +q is placed a distance r from another
charge also of +q. A repulsive force of magnitude F
is found to exist between them. Choose the correct
answer that describes the changes, if any, that will
occur to F:
a If one of the charges is doubled to +2q, F will halve/
double/quadruple/quarter and repel/attract.
b If both charges are doubled to +2q, F will halve/
double/quadruple/quarter and repel/attract.
c If one of the charges is changed to −2q, F will halve/
double/quadruple/quarter and repel/attract.
d If the distance between the charges is halved to
0.5r, F will halve/double/quadruple/quarter and
repel/attract.
Action
positive/negative attraction/repulsion
negative
4
attraction/repulsion
attraction/repulsion
A point charge, q, is moved from a position 30 cm
away from a test charge to a position 15 cm from the
same test charge. If the magnitude of the original
electric field, E, was 6.0 × 103 N C−1, what is the
magnitude of the electric field at the new position?
A 3.0 × 103 N C−1
B 6.0 × 103 N C−1
C 12.0 × 103 N C−1
D 24.0 × 103 N C−1
A hydrogen atom consists of a proton and an
electron separated by a distance of 53 pm
(picometres: 1 pm = 1 × 10−12 m). Calculate the
magnitude and sign of the force applied to a proton
carrying a charge of +1.602 × 10−19 C by an electron
carrying a charge of −1.602 × 10−19 C.
(Use ε0 = 8.8542 × 10−12 C2 N−1 m−2.)
MODULE 4 | ELECTRICITY AND MAGNETISM
Chapter review
KEY TERMS
charge
conductor
coulomb
dipole
electric field
electric field strength
electrical potential
electron
electrostatic force
elementary charge
field
field line
insulator
ion
ionised
metal
net charge
neutron
non-metal
point charge
potential difference
proton
REVIEW QUESTIONS
1
Which of the following materials is a good insulator?
A copper
B silicon
C porcelain
D aluminium
2
Approximately how many electrons make up a charge
of −3 C?
3
What will be the approximate charge on 4.2 × 1019
protons?
4
An alpha particle consists of two protons and two
neutrons. Calculate the charge on an alpha particle.
5
Which of the following materials is considered a
semiconductor?
A copper
B silicon
C porcelain
D aluminium
6
7
Which of the following statements is correct when
drawing electric field lines around a charged object?
A Electric field lines go from negatively charged
objects to positively charged objects.
B Electric field lines start and end at 90° to the
surface, with no gap between the lines and the
surface.
C Field lines can cross.
D The closer together field lines are, the weaker the
strength of the field.
Calculate the magnitude of the force applied to an oil
drop carrying a charge of 3.00 mC in a uniform electric
field of 7.50 N C−1.
8
Explain the difference between electrical potential and
potential difference.
9
Calculate the potential difference that exists between
two points separated by 25.0 mm, parallel to the field
lines, in an electric field of strength 1000 V m−1.
10 Where will the electrical field strength be at a
maximum, between two plates forming a uniform
electric field?
A close to the positive plate
B close to the earthed plate
C at all points between the plates
D at the mid-point between the plates
11 Choose the correct terms from those shown in bold
type to complete the relationship between work done
and potential difference:
When a positively charged particle moves across a
potential difference from a positive plate towards
an earthed plate, work is done by the field/charged
particle on the field/charged particle.
12 Calculate the work done to move a positively charged
particle of 2.5 × 10−18 C a distance of 3.0 mm towards a
positive plate in a uniform electric field of 556 N C−1.
13 A particular electron gun accelerates an electron a
distance of 12 cm between a pair of charged plates
across a potential difference of 15 kV. What is the
magnitude of the force acting on the electron? (Use
qe = −1.6 × 10−19 C.)
14 A researcher sees a stationary oil drop with a mass
of 1.161 × 10−14 kg between two horizontal parallel
plates. Between the plates there is an electric field of
strength 3.55 × 104 N C−1. The field is pointing vertically
downwards. Calculate how many extra electrons are
present on the oil drop. (Use qe = −1.602 × 10−19 C and
g = 9.8 N kg−1.)
15 For each of the following charged objects in a uniform
electric field, determine if work was done on the field,
by the field or if no work is done.
a An electron moves towards a positive plate.
b A positively charged point remains stationary.
c A proton moves towards a positive plate.
d A lithium ion (Li+) moves parallel to the plates.
e An alpha particle moves away from a negative plate.
f A positron moves away from a positive plate.
CHAPTER 12 | ELECTROSTATICS
357
CHAPTER REVIEW CONTINUED
16 An alpha particle is located in a parallel plate
arrangement that has a uniform electric field of
34.0 V m−1.
a Calculate the work done to move the alpha
particle a distance of 1.00 cm from the earthed
plate to the plate with a positive potential.
(qα = +3.204 × 10−19 C.)
b For the situation in part a, decide whether work
was done on the field, by the field or if no work
was done.
17 A charge of +q is placed a distance r from another
charge also of +q. A repulsive force of magnitude F
is found to exist between them. Choose the correct
options from those in bold type to describe the
changes, if any, that will occur to the force in the
following situations.
a The distance between the charges is doubled to 2r,
so the force will halve/double/quadruple/quarter
and repel/attract.
b The distance between the charges is halved to 0.5r,
so the force will halve/double/quadruple/quarter
and repel/attract.
c The distance between the charges is doubled
and one of the charges is changed to −2q, so the
force will halve/double/quadruple/quarter and
repel/attract.
18 A gold(III) ion is accelerated by the electric field
created between two parallel plates separated by
0.020 m. The ion carries a charge of 3qe and has
a mass of 3.27 × 10−25 kg. A potential difference of
1000 V is applied across the plates. The work done
to move the ion from one plate to the other results in
an increase in the kinetic energy of the gold(III) ion. If
the ion starts from rest, calculate its final speed. (Use
qe = −1.602 × 10−19 C.)
358
MODULE 4 | ELECTRICITY AND MAGNETISM
19 Calculate the force that would exist between two point
charges of 5.00 mC and 4.00 nC separated by 2.00 m.
(Use k = 9.0 × 109 N m2 C−2.)
20 A point charge of 2.25 mC is positioned on top of
an insulated rod on a table. At what distance above
the point charge should a sphere of mass 3.00 kg
containing a charge of 3.05 mC be located, so that it is
suspended in the air? (Use k = 9 × 109 N m2 C−2.)
21 A charged plastic ball of mass 5.00 g is placed in a
uniform electric field pointing vertically upwards with
a strength of 300.0 N C−1. Calculate the magnitude
and sign of the charge required on the ball in order to
create a force upwards that exactly equals the weight
force of the ball.
22 Calculate the repulsive force on each proton in a
helium nucleus separated in a vacuum by a distance
of 2.50 fm. (Use k = 9 × 109 N m2 C−2; 1 fm = 1 × 10−15 m;
and qp = +1.602 × 10−19 C.)
23 Two point charges (30.0 cm apart in air) are charged
by transferring electrons from one point to another.
Calculate how many electrons must be transferred
so that an attractive force of 1.0 N exists. Consider
only the magnitude of qe in your calculations.
(Use k = 9 × 109 N m2 C−2 and qe = −1.602 × 10−19 C.)
24 After completing the activity on page 338, reflect on
the inquiry question: How do charged objects interact
with other charged objects and with neutral objects? In
your response, discuss the electrostatic force and how
it works.
CHAPTER
Electric circuits
Every object around you is made up of charged particles. When these particles
move relative to one another, we experience a phenomenon known as ‘electricity’.
This chapter looks at fundamental concepts such as current and voltage that
scientists have developed to explain electrical phenomena, and at electric circuits
(the basis of much of our modern technology). This chapter also introduces a
range of circuits, from simple series circuits to the complex parallel wiring systems
that make up a modern home, in addition to the power and work of circuits (using
Ohm’s law). This will provide the foundation for studying practical electrical circuits.
Content
INQUIRY QUESTION
How do the processes of the transfer and the transformation
of energy occur in electric circuits?
By the end of this chapter you will be able to:
• investigate the flow of electric current in metals and apply models to represent
current, including:
-- I =
q
t
(ACSPH038) CCT ICT N
• investigate quantitatively the current–voltage relationships in ohmic and nonohmic resistors to explore the usefulness and limitations of Ohm’s law using:
-- V =
W
q
-- R =
V
I
(ACSPH003, ACSPH041, ACSPH043) ICT N
• investigate quantitatively and analyse the rate of conversion of electrical
energy in components of electric circuits, including the production of heat
and light, by applying P = VI and E = Pt and variations that involve Ohm’s law
(ACSPH042) ICT N
• investigate qualitatively and quantitatively series and parallel circuits to relate the
flow of current through the individual components, the potential differences across
those components and the rate of energy conversion by the components to the
laws of conservation of charge and energy, by deriving the following relationships:
(ACSPH038, ACSPH039, ACSPH044) ICT N
-- ∑ I = 0 (Kirchhoff’s current law – conservation of charge)
-- ∑ V = 0 (Kirchhoff’s voltage law – conservation of energy)
-- Rseries = R1 + R2 + … + Rn
--
1
Rparallel
=
1
R1
+
1
R2
+…+
1
Rn
• investigate quantitatively the application of the law of conservation of energy to
the heating effects of electric currents, including the application of P = VI and
variations of this involving Ohm’s law (ACSPH043) CCT N
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
13.1 Electric current and circuits
A flow of electric charge is called electric current. Current can be carried by
moving electrons in a wire or by ions in a solution. This section explores current as
it flows through wires in electric circuits.
Electric circuits are involved in much of the technology used every day and are
responsible for many familiar sights (Figure 13.1.1). To construct electric circuits,
you must know about the components of a circuit and be able to read circuit diagrams.
FIGURE 13.1.1
Electric circuits are responsible for lighting up whole cities.
ELECTRIC CIRCUITS
–
An electric circuit is a path made of conductive material, through which charges
can flow in a closed loop. This flow of charges is called electric current. The most
common conductors used in circuits are metals, such as copper wire. The charges
that flow around the circuit within the wire are negatively charged electrons. The
movement of electrons in the wire is called electron flow.
A simple example of an electric circuit is shown in Figure 13.1.2. The light bulb
is in contact with the positive terminal (end) of the battery; a copper wire joins the
negative terminal of the battery to one end of the filament in the light bulb. This
arrangement forms a closed loop that allows electrons within the circuit to flow
from the negative terminal towards the positive terminal of the battery. The battery
is a source of energy. The light bulb converts (changes) this energy into other forms
of energy, such as heat and light, when the circuit is connected.
If a switch is added to the circuit in Figure 13.1.2, the light bulb can be turned
off and on. When the switch is closed, the circuit is complete. The current flows in a
loop along a path made by the conductors and then returns to the battery.
When the switch is open, there is a break in the circuit and the current can no
longer flow. This is what happens when you turn off the switch for a lamp or TV. A
circuit where the conducting path is broken is often called an open circuit.
A switch on a power point or an appliance allows you to break the circuit. A
break in the circuit occurs when two conductors in the switch are no longer in
contact. This stops the flow of current and the appliance will not work.
FIGURE 13.1.2 When there is a complete
conduction path from the positive terminal of a
battery to the negative terminal, a current flows.
Current will flow in a circuit only when the circuit forms a continuous (closed)
loop from one terminal of a power supply to the other terminal.
filament
connecting
wires
insulator
+
360
external
connections
MODULE 4 | ELECTRICITY AND MAGNETISM
REPRESENTING ELECTRIC CIRCUITS
A number of different components can be added to a circuit. It is not necessary to
be able to draw detailed pictures of these components; simple symbols are much
clearer. The common symbols used to represent the electrical components in electric
circuits are shown in Figure 13.1.3.
Device
Symbol
Device
wires crossed
not joined
Symbol
cell
(DC supply)
battery of cells
(DC supply)
wires joined,
junction of conductor
fixed resistor
AC supply
or
light bulb
or
ammeter
A
voltmeter
V
or
fuse
diode
switch open
earth or ground
FIGURE 13.1.3
switch closed
Some commonly used electrical devices and their symbols.
Circuit diagrams
When building anything, it is important that the builder has a clear set of instructions
from the designer. This is as much the case for electric circuits as it is for a tall
building or a motor vehicle.
Circuit diagrams are used to clearly show how the components of an electric
circuit are connected. They simplify the physical layout of the circuit into a diagram
that is recognisable by anyone who knows how to interpret it. You can use the list
of common symbols for electrical components (Figure 13.1.3) to interpret any
circuit diagrams.
The circuit diagram in Figure 13.1.4b shows how the components of the torch
shown in Figure 13.1.4a are connected in a circuit. The circuit can be traced by
following the straight lines representing the connecting wires.
(a)
switch
contact
positive
terminal
(b)
on
negative
terminal
FIGURE 13.1.4 (a) A battery and light bulb connected by conductors in a torch constitute an electric
circuit. (b) The torch’s circuit can be represented by a simple circuit diagram.
CHAPTER 13 | ELECTRIC CIRCUITS
361
Conventional current vs. electron flow
I
+
–
electron
flow
When electric currents were first studied, it was (incorrectly) thought the charges
that flowed in circuits were positive. Because of this, scientists traditionally talked
about electric current as if current flowed from the positive terminal of the battery
to the negative terminal. This convention is still used today, even though we know
now that it is actually the negative charges (electrons) that flow around a circuit, in
the opposite direction.
On a circuit diagram current is indicated by a small arrow and the symbol I. This
is called conventional current or simply current. The direction of conventional
current is opposite to the direction of electron flow (Figure 13.1.5).
Conventional current (or current), I, flows from the positive terminal of a power
supply to the negative terminal.
Electron flow (or electron current) refers to the flow of electrons from the
negative terminal to the positive terminal of a power supply.
Conventional current (I) and
electron flow are in opposite directions. The long
terminal of the battery is positive.
FIGURE 13.1.5
QUANTIFYING CURRENT
As discussed in Chapter 12, electrons move easily through conductors, as the
electrons are only very slightly attracted to their respective nuclei. For this reason,
wiring in electric circuits needs to consist of a good conductor such as a metal.
Copper is an example of a very good conductor, which is why it is often used in
telecommunications and electrical and electronic products.
Charges are not used up or lost when a current flows around a circuit. The
charge carriers (electrons) are conserved at all points in a circuit.
As current flows, electrons travel into the wire at the negative terminal of the
battery. As electrons flow around a circuit, they remain within the metal conductor.
They flow through the circuit and return to the battery at the positive terminal but
are not lost in between.
In common electrical circuits, a current consists of electrons flowing within
a copper wire (Figure 13.1.6). This current, I, can be defined as the amount of
charge that passes through a point in the conductor per second.
copper wire
+
–
electron
The number of electrons that pass through a point per second gives a measure of the
current. Because the electrons do not leave the wire, current is conserved in all parts of the circuit.
FIGURE 13.1.6
An equation to express this is:
I = qt
where:
I is the current (in A)
q is the amount of charge (in C)
t is the time passed (in s).
Current is measured in amperes, or amps (A). One ampere is equivalent to one
coulomb per second (C s−1).
362
MODULE 4 | ELECTRICITY AND MAGNETISM
Current is a flow of electrons. It is equal to the number of electrons (ne) that flow
through a particular point in the circuit multiplied by the charge on one electron
(qe = −1.6 × 10−19 C) divided by the time that has elapsed in seconds (t). This makes
the equation:
I = qt =
neqe
t
.
Note that a flow of 1 C of positive charge to the right in 1 second is equivalent
to a flow of 1 C of negative charge to the left in 1 second. That is, both situations
represent a conventional current of 1 A to the right.
A typical current in a circuit powering a small DC motor is about 50 mA. Even
with this seemingly small current, approximately 3 × 1017 electrons flow past any
point on the wire each second.
Worked example 13.1.1
QUANTIFYING CURRENT
Calculate the number of electrons that flow past a particular point each second
in a copper wire that carries a current of 0.5 A.
Thinking
Rearrange the equation I =
q the subject.
Working
q
t
to make
I=
I×t =
q
t
( )×t
q
t
∴q = I ×t
Calculate the amount of charge that
flows past the point in question by
substituting the values given.
q = 0.5 × 1
Find the number of electrons by
dividing the charge by the charge of
an electron (1.6 × 10−19 C).
ne =
= 0.5 C
=
q
qe
0.5
1.6×10−19
= 3.13 × 1018 electrons.
Worked example: Try yourself 13.1.1
QUANTIFYING CURRENT
FIGURE 13.1.7 A digital ammeter (labelled with
an A) measures current in a circuit.
Calculate the number of electrons that flow past a particular point each second
in a copper wire that carries a current of 0.75 A.
MEASURING CURRENT: THE AMMETER
Current is commonly measured by a device called an ammeter (Figure 13.1.7).
Figure 13.1.7 shows the ammeter connected along the same path taken by the
current flowing through the light bulb. This is referred to as connecting the ammeter
‘in series’. Series circuits are covered in more detail in Section 13.4. The positive
terminal of the ammeter is connected so that it is closest to the positive terminal of
the power supply. The negative terminal of the ammeter is closest to the negative
terminal of the power supply.
Measuring the current is possible because charge is conserved at all points in
a circuit. This means that the current that flows into a light bulb is the same as the
current that flows out of the light bulb. An ammeter can therefore be connected
before or after the bulb in series to measure the current. Table 13.1.1 lists some
typical values for electric current in common situations.
TABLE 13.1.1
Typical values for electric current.
Situation
Current
Lightning
10 000 A
Starter motor in car
200 A
Fan heater
10 A
Toaster
3A
Light bulb
400 mA
Pocket calculator
5 mA
Nerve fibres in body
1 μA
CHAPTER 13 | ELECTRIC CIRCUITS
363
ANALOGIES FOR ELECTRIC CURRENT
Since we cannot see the movement of electrons in a wire, it is sometimes helpful to
use analogies or ‘models’ to visualise or explain the way an electric current behaves.
It is important to remember that no analogy is perfect: it is only a representation and
there will be situations where the electric current does not act as you would expect
from the analogy.
Water model
A very common model is to think of electric charges as water being pumped around
a pipe system, as shown in Figure 13.1.8. The battery pushes electrons through
the wires just like a pump pushes water through the pipes. Since water cannot be
compressed, the same amount of water flows in every part of a pipe, just as the
electric current is the same in every part of a wire.
water-driven
turbine
pump
light bulb
volts
DC voltage source
water
flow
electron
flow
Comparing a DC circuit to the flow of water
upper water
tank
lower water
tank
FIGURE 13.1.8
PHYSICSFILE
ICT
Lightning power
Nature provides the most extreme
examples of electric currents in the
form of lightning. The movement of
a huge number of charges heats a
channel of air up to 30 000°C. The air
then ionises and a current can flow.
Some lightning bolts have currents
greater than 200 000 A.
It is estimated that at any one time
there are 2000 lightning storms around
the globe. These create more than
100 lightning strikes every second.
The energy from just one large
thunderstorm would be enough to
power all the homes in Australia for a
few hours.
364
An electric current can be compared to water flowing through a pipe system.
Light bulbs in an electric circuit are like turbines: whereas the turbine converts
the gravitational energy of the water into kinetic energy, a light bulb converts
electrical energy into heat and light. The water that has flowed through the turbine
flows back to the pump that provides the energy needed for it to keep flowing. This
model explains the energy within a circuit quite well.
• The power supply transfers energy to the electrons and so the electrons gain
potential energy.
• The energy of the electrons is converted into other forms when the electrons
pass through the components in the circuit.
One of the limitations of the water model is that you usually cannot see water
moving through a pipe and so you have to imagine what is happening in the pipe
and then compare it to the motion of electrons in the wires.
Bicycle chain model
Although electrons move relatively slowly through a conductor, electric effects are
almost instantaneous. For example, the delay between flicking a light switch and
the light coming on is too small to be noticed. One way of understanding this is to
compare an electric current to a bicycle chain (Figure 13.1.9).
MODULE 4 | ELECTRICITY AND MAGNETISM
switch
PHYSICSFILE
ICT
Electron flow
+
battery
motion of chain
–
light bulb
wire
motion of electrons
FIGURE 13.1.9 Electrons in a wire are like the links of a bicycle chain. Just like the links of a bicycle
chain, electrons move together in a conductor.
Since a wire is full of electrons that all repel each other, moving one electron
affects all the others around it. An electric current is like a bicycle chain: even if the
cyclist pedals slowly, the links in the chain mean that energy is instantly transferred
from the pedals to the wheel.
In this model, the pedals of the bicycle are like the battery of the electric circuit:
the pedals provide the energy which causes the chain to move. This model reinforces
a number of important characteristics of an electric current.
• Electric effects are nearly instantaneous, just as there is no delay between turning
the pedals and the back wheel of the bicycle turning.
• Charges in an electric current are not ‘consumed’ or ‘used up’, just as links in the
chain are not used up.
• The amount of energy provided by an electric current is not entirely dependent
on the current. This is like when a cyclist changes gears to give the same amount
of energy to the bicycle while pedalling at different rates.
Although the bicycle chain model can be a helpful analogy, there are a number
of important differences between a bicycle chain and an electric circuit.
• The number of charges flowing in an electric current is much larger than the
number of links in a bicycle chain.
• Electrons in a wire do not touch one another like the links in a chain.
In any piece of conducting material,
such as copper wire, electrons are
present throughout the material. If
there is no current flowing, this means
there is no net flow of electrons, but the
electrons are still present.
When you connect a piece of
conducting material to the negative
terminal of a battery, the negative
terminal tries to ‘push’ the electrons
away. However, the electrons will
not flow if the circuit is open. This is
because the electrons at the open part
of the circuit have effectively reached
a dead end, like cars stopped at a
road block. This prevents all the other
electrons in the material from flowing,
like a long traffic jam caused by the
road block. When you close the circuit,
you create a clear pathway for the
electrons to flow through. This means
the electron closest to the negative
terminal forces the next electron to
move, and so on, all the way around
the circuit. Therefore all electrons move
almost simultaneously throughout
the circuit so that electronic devices,
such as light bulbs, seem to turn on
immediately after you flick the switch.
CHAPTER 13 | ELECTRIC CIRCUITS
365
13.1 Review
SUMMARY
• Electrons move easily through conductors, so
wiring in electric circuits needs to consist of a
good conductor such as a metal.
• Current, I, is defined as the amount of charge, q,
that passes through a point in a conducting wire
per second. It has the unit amperes or amps (A),
which are equivalent to coulombs per second. The
equation for current is:
• Current will flow in a circuit only when the circuit
forms a continuous (closed) loop from one
terminal of a power supply to the other terminal.
I=
• When an electric current flows, electrons all around
the circuit move towards the positive terminal, at
the same time. This is called electron flow.
q
t
=
neqe
t
.
• Current is measured with an ammeter connected
along the same path as the current flowing (that
is, in series) within the circuit.
• Conventional current in a circuit flows from the
positive terminal to the negative terminal.
KEY QUESTIONS
1
What are the requirements for current to flow in a
circuit?
2
List the electronic devices shown in the circuit diagram
below.
5
A car headlight may draw a current of 5 A. How much
charge will have flowed through it in:
a 1 second?
b 1 minute?
c 1 hour?
6
Using the values given in Table 13.1.1 on page 363,
find the amount of charge that would flow through a:
a pocket calculator in 10 min
b car starter motor in 5 s
c light bulb in 1 h.
7
1 × 1020 electrons flow past a point in a copper wire in
4 seconds. Calculate:
a the amount of charge, in coulombs, that moves past
a point in this time
b the current, in amps.
8
3.2 C flow past a point in a copper wire in 10 seconds.
Calculate:
a the number of electrons that move past a point in
this time
b the current, in amps.
A
3
4
366
Why do scientists refer to conventional current as
flowing from positive to negative?
A Protons flow from the positive terminal of a battery
to the negative terminal.
B Electrons flow from the positive terminal of a battery
to the negative terminal.
C Originally, scientists thought charge carriers were
positive.
D Charges flow in both directions in a wire;
conventional current refers to just one of the flows.
A charge of 30 C flows through a copper wire. What
current flows through the wire in:
a 10 seconds?
b 1 minute?
c 1 hour?
MODULE 4 | ELECTRICITY AND MAGNETISM
13.2 Energy in electric circuits
Electrons don’t move around a circuit unless they are given energy. This energy
can be given by a battery (Figure 13.2.1). This is because inside every battery, a
chemical reaction is taking place. The chemical reactions provide potential energy
to the electrons inside it. When a circuit connects two ends of the battery, the
potential energy of the electrons is converted into kinetic energy and the electrons
move through the wire.
FIGURE 13.2.1
Chemical energy is stored within batteries.
Chemical reactions in the battery drive electrons towards its negative terminal.
The electrons at the negative terminal repel each other. This repulsion moves them
into the wire. At the positive terminal, electrons in the wire are attracted to the
positive charges created by the deficiency of electrons. This attraction causes them
to move into the battery. The net effect of electrons flowing into the wire at one end
and out at the other end is that an electric current flows through the wire.
ENERGY IN CIRCUITS
Chemical energy stored inside a battery is transformed (changed) into electrical
potential energy. This potential energy is stored as a separation of charge between
the two terminals of the battery. This can be visualised as a ‘concentration’ of charge
at either end of the battery. One terminal (the negative terminal) has a concentration
of negative charges; the other terminal (the positive terminal) has a concentration of
positive charges. Once the battery is connected within a device, chemical reactions
will, for some time, maintain this difference in charge between the two terminals.
The difference in charge between the two terminals of a battery can be quantified
(given a numerical value) as a difference in the electrical potential energy per unit
charge. This is commonly called potential difference (V ) and is measured in
volts (V).
It is this potential difference at the terminals of the battery that provides the
energy to a circuit.The energy is then transferred (passed) to different components
in the circuit. At each component the energy is transformed into a different type of
energy. For example, the energy is transformed into heat and light if the component
is a light bulb. If the component is a fan, the energy is transformed into motion
(kinetic energy) and some heat and sound.
CHAPTER 13 | ELECTRIC CIRCUITS
367
Cells and batteries
A single cell generates electricity by converting chemical energy to electrical
potential energy. If a series of cells are added together, it is called a battery. Often a
series of cells are packaged in a way that makes it look like a single device, but inside
is a battery of cells connected together (Figure 13.2.2). The terms ‘battery’ and ‘cell’
can be used interchangeably as the term ‘battery’ is frequently used in common
language to describe a cell.
FIGURE 13.2.2 A mobile-phone battery. The term ‘battery’ actually refers to a group of electric cells
connected together.
+ +
+ +
– –
– –
FIGURE 13.2.3 A potential difference exists
between these two objects due to the difference
in charge concentration. Electrons flow from the
negative object to the positive object, as shown
by the arrow, until the potential difference is zero.
Conductors and potential difference
If we use a conductor to link two bodies between which there is a potential difference,
charges will flow through the conductor until the potential difference is equal to
zero (Figure 13.2.3). For the same reasons, when a conductor is charged, charges
will move through it until the potential difference between any two points in the
conductor is equal to zero.
Energy transfers and transformations in a torch
FIGURE 13.2.4 An X-ray image of the internal
structure of a torch. The bulb and two batteries
are clearly visible.
368
A torch is a simple example of how energy is transformed and transferred within
a circuit. In the torch shown in Figure 13.2.4, chemical energy in the battery is
transformed to electrical potential energy. There are two batteries connected so a
bigger potential difference is available. Energy can be transferred to the light bulb
once the end terminals of the batteries have been connected to the torch’s circuit:
that is, when the torch is switched on.
The electrical potential difference between the battery’s terminals causes
electrons within the circuit to move. The electrons flow through the wires of the
torch. These electrons collide with the atoms within the small wire (filament) in the
torch’s light bulb and transfer kinetic energy to them. This transfer of kinetic energy
means that the particles inside the filament move faster and faster and the filament
gets very hot. When hot, the filament emits visible light.
The energy changes can be summarised as:
transformed
→ electrical potential energy
chemical energy 
transformed
→ kinetic energy (electrons)
electrical potential energy 
transformed
→ kinetic energy (filament atoms)
kinetic energy (electrons) 
transformed
→ thermal energy + light.
kinetic energy (filament atoms) 
MODULE 4 | ELECTRICITY AND MAGNETISM
Eventually, when most of the chemicals within the battery have reacted, the
battery is no longer able to provide enough potential difference to power the torch.
This is because the chemical reaction has slowed and electrons are not being driven
to the negative terminal. The torch stops working and the batteries are said to have
gone flat.
Similar energy transfers and transformations take place every time electrical
energy is used.
EXPLAINING POTENTIAL DIFFERENCE
When charges are separated in a battery, each charge gains electrical potential
energy. In a similar way, if a mass is lifted above the ground it gains gravitational
potential energy. The change in electrical potential energy of each charge is known
as the potential difference (V ).
As you can see in Figure 13.2.5, when you lift an object to some height above the
ground, you have done some work and have placed it in the field where it has more
gravitational potential energy (U ) available to it.
PHYSICSFILE
L
Volts and voltage
Somewhat confusingly, scientists use
the symbol ‘V’ for both the quantity
potential difference and its unit of
measurement, the volt. For the quantity
potential difference, we use italics: V.
For the unit volts, the symbol is not in
italics: V. The context usually makes it
clear which meaning is intended.
For this reason, potential difference is
often referred to as ‘voltage’.
The changing potential energy of a
moving mass in a gravitational field
m
PHYSICSFILE
B
ICT
Birds on a wire
g⃗
A
UB > UA
FIGURE 13.2.5 Lifting a mass above ground level increases its gravitational potential energy just as
moving an electron to the negative terminal of an electric cell gives it more electrical potential energy.
The term ‘potential difference’ can often cause confusion. Stated simply, it
means that there is a difference in electrical potential energy per charge between
two points. The potential difference of a battery refers to the difference in electrical
potential energy per charge at either terminal of the battery. You may see potential
difference used to describe what happens on either side of a component, such as a
light globe. If you measure the potential energy of charges on one side of a globe
and compare it to the potential energy of the charges on the other side, you would
find that there is a difference. The difference equates to the energy transferred from
the charges to the globe to light it up.
Birds can sit on power lines and not get
electrocuted, even though the wires are
not insulated (Figure 13.2.6).
For a current to flow through a bird on a
wire, there would have to be a potential
difference between its two feet. Since
the bird has both feet touching the
same wire, which might be at a very
high potential (voltage), there is no
potential difference between the bird’s
feet. If the bird could stand on the wire
and touch any other object such as the
ground or another wire, then it would
get a big electric shock. This is because
there would be a potential difference
between the wire and the other object,
and current would flow.
Quantifying potential difference: Voltage
As with other forms of energy, it is useful to be able to quantify the amount of potential
difference in a given situation. Potential difference is formally defined as the amount
of electrical potential energy given to each coulomb of charge. As an equation, it is:
V = Wq = ∆qU
where:
V is potential difference (in V)
W is the work, or electric potential energy (in J)
q is charge (in C).
FIGURE 13.2.6 There is no potential
difference between each bird’s feet.
CHAPTER 13 | ELECTRIC CIRCUITS
369
Since work is measured in joules and charge in coulombs, the potential difference
is measured in joules per coulomb (J C−1). Note that the work is equal to the electric
potential energy. This quantity has been assigned a unit, the volt (V). So a potential
difference of 1 J C−1 is equal to 1V. When a battery is labelled 9V, this means that the
battery provides 9 J of energy to each coulomb of charge.
Worked example 13.2.1
DEFINITION OF POTENTIAL DIFFERENCE
Calculate the amount of electrical potential energy (work) carried by 5 C of
charge at a potential difference of 10 V.
Thinking
Working
Recall the definition of potential
difference.
V = Wq
Rearrange this to make work the subject.
W = Vq
Substitute the appropriate values
and solve.
W = 10 × 5
= 50 J
Worked example: Try yourself 13.2.1
DEFINITION OF POTENTIAL DIFFERENCE
A car battery can provide 3600 C charge at 12 V.
How much electrical potential energy (work) is stored in the battery?
Measuring voltage: The voltmeter
Voltage (potential difference) is usually measured by a device called a voltmeter.
Voltmeters are wired into circuits differently from ammeters. Unlike an ammeter,
which measures the current passing through a wire, a voltmeter measures the change
in voltage (potential difference) as current passes through a section of the circuit
(e.g. a component or combination of components). This means that one wire of the
voltmeter is connected to the circuit before the component and the other wire is
connected to the circuit after the component. This is called connecting the voltmeter
‘in parallel’, making a parallel circuit.
In Figure 13.2.7, the voltmeter is connected to the circuit either side of the light
globe, that is, in parallel. This is so that it can measure the voltage drop (potential
difference) across the light globe. Like the ammeter, it is important to connect the
voltmeter with the positive terminal closest to the positive terminal of the power
supply. The voltmeter’s negative terminal is connected closest to the negative
terminal of the power supply.
FIGURE 13.2.7
370
MODULE 4 | ELECTRICITY AND MAGNETISM
A voltmeter measures the voltage change (in this case, 6.23 V) across a light globe.
QUANTIFYING ELECTRICAL ENERGY
Work done by a circuit
In electrical circuits, electrical potential energy is converted into other forms of
energy. When energy is changed from one form to another, work is done. (Work is
covered in more detail in Chapter 5.) The amount of energy provided by a particular
circuit can be calculated using the definitions for potential difference and current:
GO TO ➤ Section 5.2, page 155
V = Wq
I = qt
Rearranging the definition of voltage gives:
W = Vq
Using the definition of current:
q = It
Therefore:
W = E = VIt
where:
W is the work done (in J). This is the same as the energy provided by the
current (E), also measured in J.
V is the potential difference (in V)
I is the current (in A)
t is the time (in s).
This gives us a practical way to calculate the energy used in a circuit from
measurements we can make.
Worked example 13.2.2
USING E = VIt
A potential difference of 12 V is used to generate a current of 750 mA to heat
water for 5 minutes.
Calculate the energy transferred to the water in that time.
Thinking
Convert the quantities to SI units.
Working
750 mA
1000
= 0.750 A
5 min × 60 s = 300 s
Substitute values into the equation
and calculate the amount of energy in
joules.
E = VIt
= 12 × 0.750 × 300
= 2700 J
Worked example: Try yourself 13.2.2
USING E = VIt
A potential difference of 12 V is used to generate a current of 1750 mA to heat
water for 7.5 minutes.
Calculate the energy transferred to the water in that time.
CHAPTER 13 | ELECTRIC CIRCUITS
371
PHYSICSFILE ICT WE
Multimeters
The internal circuitry of a voltmeter is
significantly different from an ammeter.
Electricians and scientists who work
with electrical circuits often find it
inconvenient to keep collections of
voltmeters and ammeters so they have
found a way to bundle the circuitry for
both meters up into a single device
known as a multimeter, shown in
Figure 13.2.8.
This is much more convenient,
because the multimeter can be quickly
switched from being an ammeter to
being a voltmeter as needed. However,
when a multimeter is switched from
one mode to another, it is important to
make a corresponding change to the
way it is connected to the circuit being
measured. An ammeter is connected
in series and a voltmeter in parallel. In
fact, if a multimeter is working as an
ammeter and it is connected in parallel
like a voltmeter, it may draw so much
current that its internal circuitry will be
burnt out and the multimeter will be
destroyed.
Rate of doing work: Power
If you wanted to buy a new kettle, you might wonder how you could determine how
quickly different kettles boil water.
Printed on all appliances is a rating for the power of that device. Power is a
measure of how fast energy is converted by the appliance. In other words, power is
the rate at which energy is transformed by the components within the device. This
can also be described as the rate at which work is done. As an equation:
P = energy transformed
= Et
time
where P is the power in joules per second (J s−1). One joule per second is
1 watt (W).
The more powerful an appliance is, the faster it can do a given amount of work.
In other words, an appliance that draws more power can do the same amount of
work in a shorter amount of time. If you want something done quickly, then you
need an appliance that has a higher power rating.
Rearranging the previous relationship:
E = VIt to
E
t
=VI
and combining this with the power expression gives you:
P = Et = VI
This expression enables us to calculate the energy transformations in a circuit by
measuring voltage and current across circuit components. The power dissipated by
those components can be calculated in watts (W).
Worked example 13.2.3
USING P = VI
An appliance running on 230 V draws a current of 4 A.
Calculate the power used by this appliance.
Thinking
Working
Identify the relationship needed to
solve the problem.
P = VI
Identify the required values from the
question, substitute and calculate.
P = VI
A digital multimeter can be
used as either an ammeter or a voltmeter.
= 230 × 4
= 920 W
FIGURE 13.2.8
Worked example: Try yourself 13.2.3
USING P = VI
An appliance running on 120 V draws a current of 6 A.
Calculate the power used by this appliance.
ANALOGIES FOR POTENTIAL DIFFERENCE
The analogies used for electric current in Section 13.1 can also be used to understand
the concept of potential difference.
In the water model in Section 13.1, potential difference is similar to the
water pressure in the pipe. If the water is pumped into a raised water tank as in
Figure 13.1.8, potential difference can also be compared to the gravitational
potential energy given to each drop of water.
372
MODULE 4 | ELECTRICITY AND MAGNETISM
In the bicycle chain analogy in Section 13.1, potential difference is related to
how hard the bicycle is being pedalled. If the cyclist is pedalling hard, this would
correspond to a high voltage in which each link in the chain is carrying a larger
amount of energy than if the cyclist was pedalling slowly.
In both analogies the overall rate of energy output—that is, the power—is
related to both the current and the potential difference. In the water analogy, the
pressure in the pipe could be very high but the rate of energy transfer will depend
on how quickly the water is flowing. Similarly, a cyclist can work at the same rate by
pedalling hard with the chain moving slowly or pedalling more easily but with the
chain moving more quickly.
13.2 Review
SUMMARY
• Electric potential difference measures the difference
in electric potential energy available per unit charge.
• In a circuit, the energy required for charge
separation can be provided by a cell or battery.
The chemical energy within the cell is transformed
into electric potential energy.
• Potential difference can be defined as the work
done to move a charge against an electric field
between two points, using the equation:
• Power is the rate at which energy is transformed
in a circuit component. It is defined and quantified
by the relationships:
W = Vq
or V = Wq
P=
E
t
= VI
KEY QUESTIONS
1
Under what conditions will charge flow between two
bodies linked with a rod? Choose the correct response
from the following options.
A The potential difference between the bodies is not
zero and the rod is made of a conducting material.
B The potential difference between the bodies is not
zero and the rod is made of an insulating material.
C The potential difference between the bodies is equal
to zero and the rod is made of a conducting material.
D The potential difference between the bodies is equal
to zero and the rod is made of an insulating material.
2
A freezer has a power rating of 460 W and it is
designed to be connected to 230 V. Calculate:
a the work performed by the freezer in 5 minutes
b the current flowing through the freezer.
3
a What is the potential of a battery that gives a charge
of 10 C?
i 40 J of energy in 1 second
ii 40 J of energy in 10 seconds
iii 20 J of energy in 10 seconds
b What current flowed in each case
4
A charge of 5 C flows from a battery through an electric
water heater and delivers 100 J of heat to the water.
What was the potential difference of the battery?
5
How much charge must have flowed through a 12 V
car battery if 2 kJ of energy was delivered to the
starter motor?
6
A light bulb in a lamp that is connected to 240 V uses
3.6 kJ of electric potential energy in one minute.
a Into what type(s) of energy has the electrical energy
been transformed?
b Calculate the power of the lamp.
c Calculate the current flowing through the lamp.
7
In comparing the electrical energy obtained from a
battery to the energy of water stored in a hydroelectric
dam in the mountains, to what could the potential
difference of the battery be likened?
8
Andy wishes to measure the current and potential
difference for a light bulb. He has set up a circuit as
shown below.
M1
M2
M3
M4
In which positions (M1, M2, M3 or M4) can he place:
a a voltmeter?
b an ammeter?
CHAPTER 13 | ELECTRIC CIRCUITS
373
13.3 Resistance
• Resistance is a measure
of how hard it is for current
to flow through a particular
material.
• Resistance is measured in
ohms (Ω).
Resistance is an important concept because it links the ideas of potential difference
and current. Resistance is a measure of how hard it is for current to flow through
a particular material. As conductors allow current to pass through easily, they are
said to have low resistance. Insulators have a high resistance because they ‘resist’ or
limit the flow of charges through them.
For a particular object or material, the amount of resistance can be quantified
(given a numerical value). This means that the performance of electrical circuits can
be studied and predicted with a high degree of confidence.
RESISTANCE TO THE FLOW OF CHARGE
PHYSICSFILE
ICT
Electron movement
Even when current is not flowing,
free electrons tend to move around a
piece of metal due to thermal effects.
The free electrons are rushing around
at random with great speed. The net
speed of an electron through a wire,
however, is quite slow. Figure 13.3.1
compares the random motion of an
electron when current is not flowing
(AB) to the motion of the electron when
current is flowing (AB'). The difference
between the two paths is only small.
However, the combined effect of
countless electrons moving together
in this way represents a significant net
movement of charge.
overall direction of
electron movement
B′
Energy is required to create and maintain an electric current. For electrons to move
from one place to another, they need to first be separated from their atoms and then
given energy to move. In some materials (i.e. conductors), the amount of energy
required for this is negligible (almost zero). In insulators, a much larger amount of
energy is required.
Once the electrons are moving through the material, energy is also required to
keep them moving at a constant speed. Consider an electron travelling through a
piece of copper wire. It is common to imagine the wire as an empty pipe or hose
through which electrons flow. However, a piece of copper wire is not empty—it is
full of copper ions. These ions are packed tightly together in a lattice arrangement.
As an electron moves through the wire, it will ‘bump’ into the ions. The electron
needs constant ‘energy boosts’ to keep it moving in the right direction. This is why
an electrical device stops working as soon as the energy source (e.g. battery) is
disconnected.
OHM’S LAW
Georg Ohm (1789–1854) discovered that if the temperature of a metal wire is kept
constant, the current flowing through it is directly proportional to the potential
difference across it: mathematically, I ∝ V. This relationship is known as Ohm’s law.
This relationship means that if the potential difference across a wire is doubled,
for example, then the current flowing through the wire must also double. If the
potential difference is tripled, then the current would also triple.
Ohm’s law is usually written as:
ΔV = I R (or just V = I R)
where:
V is the potential difference (in V)
I is current (in A)
R is the constant of proportionality called resistance, in ohms (Ω).
B
This equation can be rearranged to give a quantitative (mathematical) definition
for resistance:
R = VI
A
Path AB shows the random
motion of an electron due to thermal
effects. Path AB′ shows the path of the
same electron when an electric current is
flowing in the direction indicated.
FIGURE 13.3.1
374
If an identical voltage produces two different sizes of current when separately
connected to two light bulbs, then the resistance of the two light bulbs must differ.
A higher current would mean a lower resistance of the light bulb, according to
Ohm’s law. This is because, when a conductor provides less resistance, more current
can flow.
MODULE 4 | ELECTRICITY AND MAGNETISM
Worked example 13.3.1
USING OHM’S LAW TO CALCULATE RESISTANCE
When a potential difference of 3 V is applied across a piece of wire, 5 A of current
flows through it.
Calculate the resistance of the wire.
Thinking
Working
Ohm’s law is used to calculate
resistance.
V = IR
Rearrange the equation to find R.
R=
V
I
Substitute in the values for this
situation.
R=
3
5
= 0.6 Ω
Worked example: Try yourself 13.3.1
USING OHM’S LAW TO CALCULATE RESISTANCE
An electric bar heater draws 10 A of current when connected to a 240 V
power supply.
Calculate the resistance of the element in the heater.
PHYSICS IN ACTION
ICT
S
CC
Incandescent light bulbs
The complex relationship between electric current and
temperature is put to use in a very common application:
the incandescent light bulb (Figure 13.3.2).
An incandescent light bulb consists of a thin piece of
curled or bent wire, called a filament, in a glass bulb. The
wire is usually made of tungsten or another metal with a
high melting point.
Traditionally, most household lighting was provided by
incandescent light bulbs. However, this form of lighting is
very inefficient. Only a small amount of the energy that
(a)
goes into an incandescent light bulb is transformed into
light: more than 90% of the energy is lost as heat.
More recent inventions such as fluorescent tubes and
LEDs (light-emitting diodes) are much more efficient and
are increasingly being used as alternatives to incandescent
light bulbs. Some examples of these are shown in
Figure 13.3.2. In 2009, as a strategy to reduce carbon
dioxide emissions, the Australian Government began to
phase out the use of incandescent bulbs.
(b)
FIGURE 13.3.2 (a) An incandescent bulb produces light when its filament heats up. (b) Alternatives to incandescent light bulbs include fluorescent
tubes, fluorescent bulbs and LED bulbs.
CHAPTER 13 | ELECTRIC CIRCUITS
375
OHMIC AND NON-OHMIC CONDUCTORS
Conductors that obey Ohm’s law are known as ohmic conductors. Ohmic
conductors are usually called resistors.
An ohmic conductor can be identified by measuring the current that flows
through the conductor when different potential differences are applied across it.
Worked example 13.3.2
USING OHM’S LAW TO CALCULATE RESISTANCE, CURRENT AND POTENTIAL
DIFFERENCE
The table below shows measurements for the potential difference and
corresponding current for an ohmic conductor.
V (V)
0
2
4
V2
I (A)
0
0.25
I1
0.75
Determine the missing results, I1 and V2.
Thinking
Working
Determine the factor by which
potential difference has increased
from the second column to the third
column.
4
2
=2
The potential difference has doubled.
Apply the same factor increase to
the current in the second column,
to determine the current in the third
column (I1).
I1 = 2 × 0.25
Determine the factor by which current
has increased from the second
column to the fourth column.
0.75
0.25
= 0.50 A
=3
The current has tripled.
V2 = 3 × 2
Apply the same factor increase to
the potential difference in the second
column, to determine the potential
difference in the fourth column (V2).
= 6V
Worked example: Try yourself 13.3.2
USING OHM’S LAW TO CALCULATE RESISTANCE, CURRENT AND POTENTIAL
DIFFERENCE
Resistor
4
The table below shows measurements for the potential difference and
corresponding current for an ohmic conductor.
I (A)
3
2
V (V)
0
3
9
V2
I (A)
0
0.20
I1
0.80
Determine the missing results, I1 and V2.
1
2
4
V (V)
6
8
The data from an experiment in which the current and potential difference are
measured for a device is usually plotted on an I–V graph. If the conductor is ohmic,
this graph will be a straight line, as can be seen in Figure 13.3.3.
The resistance of the ohmic conductor (or resistor) can be found from the
gradient of the I–V graph. Ohm recognised that the gradient was equal to the
inverse of the resistance:
FIGURE 13.3.3 As the resistance of an ohmic
conductor is constant, the I–V graph is a straight
line.
376
MODULE 4 | ELECTRICITY AND MAGNETISM
1
R
=
rise
run
= 84 −− 21 = 63
∴ R = 63 = 2 Ω
However, not all conductors are ohmic. The I–V graphs for non-ohmic
conductors are not straight lines (Figure 13.3.4). Light bulbs and diodes are
examples of non-ohmic conductors.
Light bulb
Diode
3
3
I (A)
4
I (A)
4
2
2
1
1
2
FIGURE 13.3.4
4
6
V (V)
8
10
12
0.5
1.0
V (V)
1.5
The I–V graph for a non-ohmic conductor is not a straight line.
Using I–V graphs to determine resistance
The inverse of resistance is defined as the ratio I/V. For an ohmic conductor, this
value is a constant regardless of the potential difference across the conductor.
However, the resistance of a non-ohmic conductor varies. The resistance of a nonohmic conductor for a particular potential difference can be found by determining
the current flowing through the conductor at this value of potential difference.
Worked example 13.3.3
CALCULATING RESISTANCE FOR A NON-OHMIC CONDUCTOR
Calculate the resistance of the light bulb with the I–V graph shown when the
potential difference is 5.0 V.
Light bulb
4
I (A)
3
2
1
2
4
6
V (V)
8
10
12
CHAPTER 13 | ELECTRIC CIRCUITS
377
Thinking
Working
From the graph, determine the current
at the required potential difference.
At V = 5 V, I = 3 A
Substitute these values into Ohm’s law
to find the resistance.
R=
V
I
=
5
3
= 1.67 Ω
Worked example: Try yourself 13.3.3
CALCULATING RESISTANCE FOR A NON-OHMIC CONDUCTOR
A 240 V, 60 W incandescent light bulb has the I–V characteristics shown in the graph.
Calculate the resistance of the light bulb at 175 V.
Light bulb
I (mA)
200
100
100
V (V)
200
RESISTORS IN SIMPLE CIRCUITS
Ohmic resistors are often used to control the amount of current in a particular
circuit. Resistors can be manufactured to produce a relatively constant resistance
over a range of temperatures. A colour-coding system is used on resistors to explain
the amount of resistance they provide, including a percentage tolerance (precision).
Figure 13.3.5 shows a resistor that uses the colour-coding system.
Ohm’s law can be used to determine the current flowing through a resistor when
a particular potential difference is applied across it. Similarly, if the current and
resistance are known, the potential difference across the resistor can be calculated.
FIGURE 13.3.5 Common resistors are electrical devices with a known resistance. The coloured bands
indicate the resistor’s resistance and tolerance.
378
MODULE 4 | ELECTRICITY AND MAGNETISM
PHYSICSFILE ICT
Colour-coded resistors
A resistor is typically a small piece of equipment which does not allow enough room
to clearly print information about the resistor in the form of numbers. A colour-coding
system is used on many resistors to convey detailed information in a small space about
the resistance and tolerance of the resistor. Figure 13.3.6 explains how to interpret the
colour-coding system for the commonest type of resistor, which has four colour bands.
Resistor colour code
Band colour
Tolerance colour code
Value
Band colour
±%
Black
0
Brown
1
Brown
1
Red
2
Red
2
Gold
5
Orange
3
Silver
10
Yellow
4
None
20
Green
5
Blue
6
Purple
7
What this means
Grey
8
Band 1
White
9
Gold
0.1
Silver
0.01
Brown
1
Green
5
Orange
000
Band 2
Band 3 Number of zeros/multiplier
Band 4 Tolerance (±%) See below
Note that the bands are closer
to one end.
Gold
5%
Resistor is 15 000 Ω or 15 kΩ ± 5%
Red
2
Red
2
Green
00000
None
20%
Resistor is 2 200 000 Ω or 2.2 MΩ ± 20%
FIGURE 13.3.6
Yellow
4
Purple
7
Silver
×0.01
Red
2%
Resistor is 47 × 0.01 Ω or 0.47 Ω ± 2%
Brown
1
Green
5
Red
00
Gold
5%
Resistor is 1500 Ω or 1.5 kΩ ± 5%
Examples of resistor colour-coding.
CHAPTER 13 | ELECTRIC CIRCUITS
379
Worked example 13.3.4
USING OHM’S LAW TO FIND CURRENT
A 100 Ω resistor is connected to a 12 V battery.
Calculate the current (in mA) that would flow through the resistor.
Thinking
Working
Recall Ohm’s law.
V = IR
Rearrange the equation to make I the
subject.
I=
Substitute in the values for this
problem and solve.
12 = 0.12 A
I = 100
Convert the answer to the required
units.
I = 0.12 A
V
R
= 0.12 × 103 mA
= 120 mA
Worked example: Try yourself 13.3.4
USING OHM’S LAW TO FIND CURRENT
The element of a bar heater has a resistance of 25 Ω.
Calculate the current (in mA) that would flow through this element if it is
connected to a 240 V supply.
Worked example 13.3.5
USING OHM’S LAW TO FIND POTENTIAL DIFFERENCE
A current of 0.25 A flows through a 22 Ω resistor.
Calculate the voltage across the resistor.
Thinking
Working
Recall Ohm’s law.
V = IR
Substitute in the values for this
problem and solve.
V = 0.25 × 22
= 5.5 V
Worked example: Try yourself 13.3.5
USING OHM’S LAW TO FIND POTENTIAL DIFFERENCE
PA
4.3
380
WS
4.6
The globe of a torch has a resistance of 5.7 Ω when it draws 700 mA of current.
Calculate the potential difference across the globe.
MODULE 4 | ELECTRICITY AND MAGNETISM
SKILLBUILDER
N
Understanding the relationship between
data, graphs and algebraic rules
Scientists use graphs to analyse the data they collect from experiments. All
graphs tell a story. The shape of the graph shows the relationship between the
variables, and this relationship can be written algebraically and numerically.
Once the algebraic rule is known, values for one variable can be substituted
and the values for the other variable can be calculated. These values can also
be determined by reading them from the graph.
For example, when investigating how current and voltage vary across a light
bulb, the following data was collected:
Current (A)
Voltage (V)
0.14
0.6
0.22
1.0
0.33
1.4
0.47
2.0
0.66
2.8
0.80
3.6
1.20
5.0
Graphing this data produced:
Voltage vs Current for a light globe
6.0
Voltage (V)
5.0
4.0
y = 4.2x + 0.05
3.0
2.0
1.0
0.20 0.40 0.60 0.80 1.00 1.20 1.40
Current (A)
The numerical values from the experiment are listed in the table and plotted
on the graph. The algebraic relationship between the variables is given by the
equation of the line:
y = 4.2x + 0.05
The value of the y-intercept (0.05) is approximately zero, so assuming that
the y-intercept is zero, and labelling the x-axis as current and the y-axis as
voltage, the relationship can be written as:
y = 4.2 × current
Using the appropriate symbols this can also be written as: V = 4.2I.
CHAPTER 13 | ELECTRIC CIRCUITS
381
13.3 Review
SUMMARY
• Resistance is a measure of how hard it is for
current to flow through a particular material.
Resistance is measured in ohms (Ω).
• Ohm’s law describes the relationship between
current, potential difference (voltage) and resistance:
• The resistance of a material depends on its length,
cross-sectional area and temperature.
• Ohmic conductors have a constant resistance.
The resistance of non-ohmic conductors varies for
different potential differences.
V = IR
KEY QUESTIONS
1
An experiment is conducted to gather data about the
relationship between current and potential difference
for three ohmic devices, labelled A, B and C. The data
is used to plot an I–V graph for each device, as shown
in the figure below.
A
A student obtains a graph of the current–voltage
characteristics of a piece of resistance wire
(see below).
10
C
8
6
4
I
Current (A)
B
3
2
5
a For a given potential difference, list the devices in
order of highest current to lowest current.
b List the devices in order of highest resistance to
lowest resistance.
The table below shows measurements for the potential
difference and corresponding current for an ohmic
conductor.
V (V)
0
2
3
V2
I (A)
0
0.25
I1
0.60
Determine the missing results, I1 and V2.
382
15
Voltage (V)
20
25
a Explain whether this piece of wire is ohmic or nonohmic.
b What current flows in this wire at a voltage of 7.5 V?
c What is the resistance of this wire?
V
2
10
MODULE 4 | ELECTRICITY AND MAGNETISM
4
A student finds that the current through a resistor is
3.5 A when a voltage of 2.5 V is applied to it.
a What is the resistance?
b The voltage is then doubled and the current is found
to increase to 7.0 A. Is the resistor ohmic or not?
5
Rose and Rachel are trying to find the resistance of
an electrical device. They find that at 5 V it draws a
current of 200 mA and at 10 V it draws a current of
500 mA. Rose says that the resistance is 25 Ω, but
Rachel maintains that it is 20 Ω. Who is right and why?
6
Nick has an ohmic resistor to which he has applied 5 V.
He measures the current as 45 mA. He then increases
the voltage to 8 V. What current will he find now?
7
Lisa finds that when she increases the voltage
across an ohmic resistor from 6 V to 10 V the current
increases by 2 A.
a What is the resistance of this resistor?
b What current does it draw at 10 V?
8
A strange electrical device has the I–V characteristics
shown in the Figure below.
1.5
I (A)
1.0
0.5
10
V (V)
20
a Is it an ohmic or non-ohmic device? Explain.
b What current is drawn when a voltage of 10 V is
applied to it?
c What voltage would be required to double the
current drawn at 10 V?
d What is the resistance of the device at:
i 10 V?
ii 20 V?
CHAPTER 13 | ELECTRIC CIRCUITS
383
13.4 Series and parallel circuits
PHYSICS INQUIRY
N
CCT
Water circuit
How do the processes of the transfer and the transformation of energy occur
in electric circuits?
COLLECT THIS…
•
10 mm diameter clear vinyl tubing
RECORD THIS…
•
two 10 cm lengths of 7 mm diameter clear vinyl tubing
•
a 10 cm length of 5 mm diameter clear vinyl tubing
Describe how the water circuit is similar to an electric
circuit.
•
bird netting
•
straight and T tube fittings
•
one bucket, with hole near the bottom to fit the 10 mm
tubing.
•
two 1 L beakers
•
a 2 L measuring flask
•
Blu Tack
•
stopwatch
Present the results of the two circuits (series and parallel)
in a table.
REFLECT ON THIS…
How do the processes of the transfer and the
transformation of energy occur in this water circuit?
Which configuration reduces the resistance in the circuit
(or increases the total flow rate): series or parallel?
What is a significant difference between the water circuit
and an electrical circuit?
DO THIS…
1 Set up the series water circuit. Use Blu Tack to hold the
tubing to the wall.
2 Fill the bucket with 2 litres of water. Time how long it
takes the water to empty from the circuit. Once the
water has completely left the circuit, measure the
volume of the water in the beaker.
3 Repeat Step 2, changing one of the tubing resistors to
the 5 mm diameter.
4 Set up the parallel water circuit. Repeat steps 2 and 3.
bucket
10 mm tubing
7 mm tubing
series circuit
beaker
bucket
parallel circuit
beaker
When a circuit contains more than one resistor, Ohm’s law alone is not sufficient to
predict the current flowing through and the potential difference across each resistor.
Additional concepts such as Kirchhoff’s rules and the idea of equivalent resistance
can be used to analyse these complex, multi-component circuits.
No matter how complex a circuit, it can always be broken up into sections in
which the circuit elements are in series or parallel. This section investigates the
difference between these two types of circuits.
RESISTORS IN SERIES
FIGURE 13.4.1 This circuit has a resistor and
light bulb connected in series.
384
Some circuits contain more than one electrical component. When these components
are connected one after another in a continuous loop, this is called a series circuit.
Components connected in this way are said to have been connected ‘in series’. The
circuit shown in Figure 13.4.1 has a resistor and a light bulb connected in series
with an electric cell.
MODULE 4 | ELECTRICITY AND MAGNETISM
Series circuits are very easy to construct, but they have some disadvantages. As
every component is connected one after the other, each component is dependent
the others. If one component is removed or breaks down, the circuit is no longer a
closed loop and it won’t work. Figure 13.4.2 shows how removing a globe from a
series circuit interrupts the entire circuit. Due to this characteristic, series circuits
with more than one component are not commonly used in the home.
(a)
(b)
FIGURE 13.4.2 (a) All components in the circuit are intact and so the circuit is a closed loop. (b) When
one light bulb is removed, the whole circuit is interrupted.
Conservation of charge
When analysing a series circuit it is important to understand that the same amount
of current flows in every part of the circuit. Since electric charges are not created or
destroyed within an electric circuit, the current flowing out of the cell must be the
same as the current flowing through the lamp, which is also the same as the current
flowing through the resistor. This current also flows unchanged back into the cell as
shown in Figure 13.4.3.
Remember that, by convention, the current is represented as flowing from the
positive terminal of the cell to the negative terminal. Electrons move in the opposite
direction.
current
FIGURE 13.4.3 In a series circuit, the same
current flows through each component.
The current in a series circuit is the same in every part of the circuit.
Kirchhoff’s voltage law
Kirchhoff’s voltage law says that the sum of the potential differences across all
the elements around any closed circuit loop must be zero. This means that the total
potential drop around a closed circuit must be equal to the total potential gain in
the circuit. This rule adheres to the law of conservation of energy. Based on the
conservation of energy, the total amount of energy gained per unit charge (voltage
gain) must equal the total amount of energy lost per unit charge (voltage loss). For
example, if a battery provides 9V to a circuit, then the sum of all of the potential
drops across the components must add to 9V.
EMF
Kirchhoff’s voltage law says that the sum of the potential differences across all
the elements around any closed circuit loop must be zero.
Kirchhoff’s voltage law can be stated algebraically as ∑V = 0.
Figure 13.4.4 shows how the voltage provided by the battery is shared across a
resistor and a lamp. The power supply in the figure is labelled EMF. Devices that are
a source of energy for a circuit are referred to as sources of EMF or electromotive
force. A battery is another example of a source of energy for a circuit. EMF,
measured in volts (V), is another term for the work done on charges to provide a
potential difference between the terminals of the power supply.
There are a number of ways to visualise the energy changes in this circuit. One
common analogy is to think of the charges as water being pumped around an elevated
water course. The water gains potential energy as it is pumped higher, and as it
flows back down the potential energy is converted into other forms. The diagram in
V1
V2
FIGURE 13.4.4 Kirchhoff’s voltage law. In this
series circuit, the sum of the potential drops
across the resistor and the lamp (i.e. V1 + V2)
will be equal to the potential difference (EMF)
provided by the battery.
CHAPTER 13 | ELECTRIC CIRCUITS
385
potential
battery
pump
Figure 13.4.5 shows how the analogy works with the energy changes that occur in
a circuit. The battery acts as a ‘pump’ that pushes electrons up to a higher energy
level and the electrons gain potential energy. As the electrons pass down through
components in the circuit, their energy is transformed into other forms.
The change in electrical energy available to electrons can also be represented
graphically, as shown in Figure 13.4.6.
FIGURE 13.4.5 An analogy for analysing a
circuit: the battery acts as a ‘pump’ which
transfers potential energy to electrons. The
electrons lose potential energy as they flow
‘down’ through components in the circuit.
Potential energy
available to electrons
max
V1
VT
V2
FIGURE 13.4.6 The electric potential energy of an electron changes as it moves around the circuit.
Some of this energy is lost as the electrons pass through the resistor. The remaining energy is lost as
the electrons pass through the bulb. In this circuit, the bulb has more resistance than the resistor.
Equivalent series resistance
Consider the circuit in Figure 13.4.7. If the resistance of the fixed resistor is R1, the
resistance of the lamp is R2 and the current flowing through both of them is I, then
Ohm’s law gives:
V1 = I × R1
and
I
I
V2 = I × R2.
The total voltage drop across the two components is:
R2
R1
Vtotal = V1 + V2 = I R1 + I R2 = I × (R1 + R2).
This equation shows the relationship between the potential difference supplied
by the cell and the potential differences of the lamp and resistor. The last part of the
V1
V2
equation also shows that the lamp and resistor can be replaced with a single resistor,
without changing the current in the circuit. The single resistor needs to have a total
Vtotal
resistance of R1 + R2.
FIGURE 13.4.7 Using Ohm’s law, it is possible
In general, a number of individual resistors connected in series can be replaced
to show the relationship between the potential
by an equivalent effective resistance (also called the total resistance) equal to the
difference supplied by the cell, Vtotal, the current
sum of the individual resistances. Figure 13.4.8 shows how two resistors can be
flowing in the circuit, I, and the resistances of
replaced with a single one.
the two components R1 and R2.
The total resistance can be derived using Kirchoff’s voltage law, so Vtotal = V1 +
V2. This is equivalent to I Rtotal = I R1 + I R2. The total resistance can then be written
as Rtotal = R1 + R2. This is shown by the equation:
Rseries = R1 + R2 + … + Rn
where Rseries is the equivalent effective series resistance and R1, R2, …, Rn are
the individual resistances.
386
MODULE 4 | ELECTRICITY AND MAGNETISM
(a)
(b)
100 Ω
100 Ω
200 Ω
Equivalent resistances can be
used in circuit analysis to simplify
a complicated circuit diagram
so that current and potential
difference can be determined.
(a) Two 100 Ω resistors can be replaced with (b) a single 200 Ω resistor to have the
same effect in a circuit.
FIGURE 13.4.8
Worked example 13.4.1
CALCULATING AN EQUIVALENT SERIES RESISTANCE
A 100 Ω resistor is connected in series with a 690 Ω resistor and a 1.2 kΩ resistor.
Calculate the equivalent series resistance.
Thinking
Working
Recall the formula for equivalent series
resistance.
Rseries = R1 + R2 + … + Rn
Substitute in the given values for
resistance. Make sure to convert kΩ
to Ω. Solve to find the equivalent series
resistance.
Rseries = 100 + 690 + 1200 = 1990 Ω
Worked example: Try yourself 13.4.1
CALCULATING AN EQUIVALENT SERIES RESISTANCE
A string of Christmas lights consists of 20 light bulbs connected in series. Each
bulb has a resistance of 8 Ω.
Calculate the equivalent series resistance of the Christmas lights.
Worked example 13.4.2
USING EQUIVALENT SERIES RESISTANCE FOR CIRCUIT ANALYSIS
Use equivalent series resistance to calculate the current flowing in the series
circuit below and the potential difference across each resistor.
9V
100 Ω
200 Ω
300 Ω
Thinking
Working
Recall the formula for equivalent series
resistance.
Rseries = R1 + R2 + R3 + … + Rn
Find the equivalent (total) resistance in the
circuit.
Rseries = 100 + 200 + 300 = 600 Ω
CHAPTER 13 | ELECTRIC CIRCUITS
387
Use Ohm’s law to calculate the current in
the circuit. Whenever calculating current in
a series circuit, use RT and the voltage of the
power supply.
Use Ohm’s law to calculate the potential
difference across each separate resistor.
I=
V
R
=
9
600
= 0.015 A
V = IR
V1 = 0.015 A × 100 Ω = 1.5 V
V2 = 0.015 A × 200 Ω = 3.0 V
V3 = 0.015 A × 300 Ω = 4.5 V
Use Kirchhoff’s voltage law to check the
answer.
VT = V1 + V2 + V3
= 1.5 V + 3.0 V + 4.5 V
= 9.0 V
Since this is the same as the
voltage provided by the cell, the
answer is reasonable.
Worked example: Try yourself 13.4.2
USING EQUIVALENT SERIES RESISTANCE FOR CIRCUIT ANALYSIS
Use equivalent series resistance to calculate the current flowing in the series
circuit below and the potential difference across each resistor.
12 V
100 Ω
690 Ω
330 Ω
RESISTORS IN PARALLEL
switch A
switch B
lamp A
lamp B
FIGURE 13.4.9 In this parallel circuit, lamp A is
off and lamp B is on.
388
One of the disadvantages of series circuits is that if a switch is opened or a device is
disconnected, then the circuit is broken and current stops flowing. In everyday life,
we often want to switch devices on and off independently. Parallel circuits allow us
to do this.
The circuit diagram in Figure 13.4.9 shows a simple parallel circuit. Even if
switch A is open (as shown), lamp B will still light up as it is part of an unbroken
complete circuit including the battery. Similarly, if switch A is closed and switch B
is opened, current will light up lamp A and not lamp B. Alternatively, both switches
could be closed to light up both lamps or both switches could be opened to switch
both lamps off.
In a series circuit, all the components are in the same loop and therefore the
same current flows through each component. In comparison, each loop of a parallel
circuit acts like an independent circuit with its own current.
Consider again the water analogy. When water flows through pipes, it is not lost.
If the pipe splits in two, some of the water flows in one pipe, and the remaining
water will flow in the other pipe. If the two sections re-join, the water comes back
together again, just as it was before the pipe was split. The same occurs with charges
flowing in a parallel circuit. Figure 13.4.10 shows how the charges go through one
globe or the other. This means that while the current in the main part of the circuit
remains constant, in the parallel section, the current is divided between each branch.
The readings on both ammeters, A1 and A2 , will be the same.
MODULE 4 | ELECTRICITY AND MAGNETISM
A1
FIGURE 13.4.10
A2
Unlike a series circuit, charges flowing in a parallel circuit have a choice of path.
The current in the main part of a parallel circuit is the sum of the currents in
each branch of the circuit:
IT = I1 + I2 + … + In.
Unlike a series circuit, in a parallel circuit the voltage is not shared between
resistors; the voltage is the same across each branch. This is because, while the
charges take different pathways, they have the same amount of energy no matter
which path they take. An advantage of parallel circuits is that globes connected in
this way are brighter than if they were connected in series. The potential energy of
the charges is not shared between the globes.
The voltage is the same in each
branch of a parallel circuit.
Kirchhoff’s current law
Parallel circuits involve junctions where current can flow in a variety of directions.
The behaviour of current at these points is predicted by Kirchhoff’s current law.
Kirchhoff’s current law states that at any junction in an electric circuit, the total
sum of all currents flowing into that junction must equal the total sum of all currents
flowing out of that junction. This is based on the principle of conservation of charge
at a point.
Kirchhoff’s current law states that at any junction in an electric circuit, the
total sum of all currents flowing into that junction must equal the total sum of all
currents flowing out of that junction.
Kirchhoff’s current law can be stated as ∑I = 0.
This rule is just an extension of the idea of conservation of charge; that is, that
charges cannot be created or destroyed. Although the number of electrons flowing
into a junction might be very large, electrons are not created or destroyed in the
junction so the same number of electrons must flow out again. This is illustrated
in Figure 13.4.11. Kirchhoff’s current law explains how current splits in a parallel
circuit. It explains why the current in the main part of the circuit is the sum of the
currents in each parallel branch.
PHYSICSFILE ICT
Christmas lights
You may have experienced the
difference between series and parallel
circuits when looking at Christmas
lights. If the bulbs on a string of
Christmas lights are connected in
series, all the lights go out when one (or
more) of the bulbs fails. Trying to find
the problematic light bulb can be quite
difficult. If the lights are connected
in parallel, when one (or more) of
the bulbs fails, the remainder of the
Christmas lights stay lit.
2A
5A
Equivalent parallel resistance
When additional resistors are added in a series circuit, the total resistance of the circuit
increases. Additional resistance means that less current flows through the circuit.
In contrast, adding an additional resistor in parallel means that more current
flows through the circuit because another path for charges has been added. This
means that the total resistance for the circuit decreases.
3A
The current flowing into any
junction must be equal to the current flowing
out of it.
FIGURE 13.4.11
CHAPTER 13 | ELECTRIC CIRCUITS
389
Parallel circuits are more complicated than series circuits, hence the formula
used to calculate the equivalent effective (total) resistance is more complicated.
The total resistance in parallel can be derived using Kirchoff’s current law. Now
we have IT = I1 + I2. Using Ohm’s Law, this is equivalent to RVtotal = VR1 + RV2 so that the
total resistance can then be written as R 1 = R1 + R1 . This is shown by the equation:
total
1
Rparallel
1
2
= R1 + R1 + + R1
1
2
n
where Rparallel is the equivalent effective resistance and R1, R2, …, Rn are the
individual resistances.
Worked example 13.4.3
CALCULATING AN EQUIVALENT PARALLEL RESISTANCE
A 100 Ω resistor is connected in parallel with a 300 Ω resistor.
Calculate the equivalent parallel resistance.
Thinking
Working
Recall the formula for equivalent
effective resistance.
1
Rparallel
=
Substitute in the given values for
resistance.
1
Rparallel
1 + 1
= 100
300
Solve for Rparallel.
1
R1
1
Rparallel
+ R1 + + R1
2
n
1 + 1 =
= 100
300
=
3
300
1
+ 300
4
300
∴ Rparallel = 75 Ω
Worked example: Try yourself 13.4.3
CALCULATING AN EQUIVALENT PARALLEL RESISTANCE
A 20 Ω resistor is connected in parallel with a 50 Ω resistor.
Calculate the equivalent parallel resistance.
Notice that in the previous worked example, the equivalent effective resistance
was smaller than the smallest individual resistance. This is because adding a resistor
provides an additional pathway for current. Since more current flows, the resistance
of the circuit has been effectively reduced.
The effective (total) resistance of a set of resistors connected in parallel is
always smaller than the smallest resistor in the set.
In a parallel circuit:
Rtotal < Rsmallest resistor
If you consider the smallest resistor in any parallel combination—say, the 20 Ω
resistor in Worked example: Try yourself 13.4.3—the addition of the 50 Ω resistor in
parallel with it allows the current an extra pathway and therefore it is easier for the
current to flow through the combination. The effective resistance of the pair must
be less than the 20 Ω alone.
390
MODULE 4 | ELECTRICITY AND MAGNETISM
Worked example 13.4.4
USING EQUIVALENT PARALLEL RESISTANCE FOR CIRCUIT ANALYSIS
Find an equivalent parallel resistance to calculate the current flowing out of the 12 V
cell in the parallel circuit shown. Also find the current flowing through each resistor.
12 V
200 Ω
100 Ω
Thinking
Working
Recall the formula for equivalent parallel
resistance.
1
Rparallel
Substitute in the given values for resistance.
1
Rparallel
Solve for Rparallel.
=
1
R1
+ R1 + + R1
2
n
1 + 1
= 100
200
1
Rparallel
1 + 1
= 100
200
=
2
200
∴ Rparallel =
200
3
1 =
+ 200
3
200
= 67 Ω
Use Ohm’s law to calculate the current in the
circuit. To calculate I, use the voltage of the power
supply and the total resistance.
Icircuit =
V
R
= 12
= 0.18 A
67
100 Ω resistor:
Use Ohm’s law to calculate the current through
each separate resistor. Remember that the voltage I100 = V = 12 = 0.12 A
R
100
through each resistor is the same as the voltage
200 Ω resistor:
of the power supply, 12 V in this case.
12 = 0.06 A
I200 = VR = 200
Use Kirchhoff’s current law to check the answers.
Icircuit = I100 + I200
0.18 A = 0.12 A + 0.06 A
This is correct, so the
answers are reasonable.
Worked example: Try yourself 13.4.4
USING EQUIVALENT PARALLEL RESISTANCE FOR CIRCUIT ANALYSIS
Use an equivalent parallel resistance to calculate the current flowing in the
parallel circuit below and through each resistor of the circuit.
10 V
30 Ω
50 Ω
CHAPTER 13 | ELECTRIC CIRCUITS
391
COMPLEX CIRCUIT ANALYSIS
Some circuits combine elements of series wiring and parallel wiring. A general
strategy for analysing these circuits is to reduce the complex circuit to a single
equivalent resistance to determine the current drawn by the circuit. It is then
possible to step back through the process of simplification to analyse each section
of the circuit as needed.
Worked example 13.4.5
COMPLEX CIRCUIT ANALYSIS
Calculate the potential difference across and current through each resistor in the
circuit below.
R1 = 10.0 Ω
R2 = 30.0 Ω
12.0 V
R3 = 40.0 Ω
R4 = 20.0 Ω
Thinking
Find an equivalent resistance for
the parallel resistors. The effective
resistance of these should be less
than the smaller resistor, that is,
smaller than 30 Ω.
Working
1
R2−3
=
1
R2
+ R1
=
1
30.0
3
1 = 4 + 3
+ 40.0
120.0 120.0
= 17.1Ω
∴ R2−3 = 120.0
7
Find an equivalent series resistance
for the circuit, as the circuit can now
be thought of as three resistors in
series: 10.0 Ω, 17.1 Ω and 20.0 Ω.
Rseries = 10.0 + 17.1 + 20.0
Use Ohm’s law to calculate the
current in the circuit. Use the supply
voltage and total resistance to do this
calculation.
V = IR
I = VR = 12.0
47.1
Use Ohm’s law to calculate the
potential difference across each
resistor (or parallel group of
resistors) in series. (Note that the
potential difference across R2 is the
same as that across R3 as they are
in parallel.)
V = IR
= 47.1 Ω
= 0.255 A
V1 = 0.255 × 10.0 = 2.55 V
V2−3 = 0.255 × 17.1 = 4.36 V
V4 = 0.255 × 20.0 = 5.10 V
Check:
2.55 + 4.36 + 5.10 ≈ 12.0 V
(with some slight rounding error)
This confirms that Kirchhoff’s voltage
law holds for this circuit.
392
MODULE 4 | ELECTRICITY AND MAGNETISM
Use Ohm’s law where necessary to
calculate the current through each
resistor.
I1 = I4 = 0.255 A
I=
V
R
I2 =
4.36
30.0
= 0.145 A
I3 =
4.36
40.0
= 0.109 A
Check: 0.145 + 0.109 ≈ 0.255 A (with
some slight rounding error)
This confirms that Kirchhoff’s current law
holds for this section.
Worked example: Try yourself 13.4.5
COMPLEX CIRCUIT ANALYSIS
Calculate the potential difference across and the current through each resistor in
the circuit below.
R1 = 20.0 Ω
R5 = 5.0 Ω
R6 = 10.0 Ω
R7 = 30.0 Ω
100.0 V
R2 = 50.0 Ω
R3 = 25.0 Ω
R4 = 100.0 Ω
RESISTORS AND POWER
A particular combination of resistors draw different amounts of power depending
on whether the resistors are wired in series or parallel. In general, since resistors in
parallel circuits draw more current than resistors in series circuits, parallel circuits
use more power than series circuits containing the same resistors.
Recall that the equation for power is:
P=V×I
where:
P is the power (in W)
V is the voltage (in V)
I is the current (in A).
PHYSICSFILE N
Identical resistors in parallel
Where identical resistors are placed
in parallel, the total resistance of the
combination can be found by simply
dividing the value of one of the resistors
by the number of resistors.
For example, the effective total
resistance of three 12 Ω resistors
connected in parallel is 4 Ω:
Rparallel = 12 ÷ 3 = 4 Ω.
The three 12 Ω resistors in parallel
could be replaced with a single 4 Ω
resistor.
Similarly, the equivalent resistance of two
10 Ω resistors placed in parallel is 5 Ω.
CHAPTER 13 | ELECTRIC CIRCUITS
393
Worked example 13.4.6
COMPARING POWER IN SERIES AND PARALLEL CIRCUITS
Consider a 100 Ω and a 300 Ω resistor wired in parallel with a 12 V cell.
Calculate the power drawn by these resistors. Compare this to the power drawn
by the same two resistors when wired in series.
Thinking
Calculate the equivalent resistance
for the parallel circuit.
Working
1
Rparallel
=
1
R1
+ R1 + + R1
∴ Rparallel =
300
4
n
2
1 + 1 =
= 100
300
3
300
1 = 4
+ 300
300
= 75 Ω
Calculate the total current drawn by
the parallel circuit.
V = IR
I = VR = 12
= 0.16 A
75
Use the power equation to calculate
the power drawn by the parallel
circuit.
P = VI
Calculate the equivalent resistance
for the series circuit.
Rseries = R1 + R2 + … + Rn
Calculate the total current drawn by
the series circuit.
V = IR
I = VR =
Use the power equation to calculate
the power drawn by the series
circuit.
P = VI
Compare the power drawn by the
two circuits.
Pparallel
Pseries
= 12 × 0.16 = 1.92 W
= 100 + 300
= 400 Ω
12
400
= 0.03 A
= 12 × 0.03 = 0.36 W
=
1.92
0.36
= 5.33
The parallel circuit draws more than 5 times
as much power as the series circuit.
Worked example: Try yourself 13.4.6
COMPARING POWER IN SERIES AND PARALLEL CIRCUITS
Consider a 200 Ω and a 800 Ω resistor wired in parallel with a 12 V cell.
PA
4.4
394
WS
4.7
WS
4.5
Calculate the power drawn by these resistors. Compare this to the power drawn
by the same two resistors when wired in series.
MODULE 4 | ELECTRICITY AND MAGNETISM
PHYSICS IN ACTION
ICT
S
Parallel connections
All household appliances and lights are
connected in parallel. This is done for two
reasons.
Figure 13.4.12 shows a TV, air-conditioner,
heater and washing machine connected in
series. Each of these devices is designed to
operate at 240 V.
240 V mains
supply
A circuit designed like the one in Figure
13.4.12 poses many problems. Firstly, all of
the devices need to be switched on for the
circuit to operate.
Secondly, the 240 V supplied to the
circuit needs to be shared among all the
components. Each component in the circuit
would receive far less than the 240 V they
require to operate. Also, as more and more
devices are added to the circuit, the share
of the 240 V would become smaller. This
system could never be practical.
The circuit diagram in Figure 13.4.13
shows how the same devices could be
connected in parallel.
Each device in the parallel circuit receives
the same voltage, 240 V. Each device can be
independently switched on or off without
affecting the others and more devices can
be added to this system without affecting
the operation of the others. For this reason,
household wiring uses parallel connections.
washing
machine
TV
airconditioner
heater
FIGURE 13.4.12
240 V mains
supply
FIGURE 13.4.13
Household appliances connected in series.
heater
airconditioner
TV
washing
machine
Household appliances connected in parallel.
CHAPTER 13 | ELECTRIC CIRCUITS
395
13.4 Review
SUMMARY
-- equivalent effective resistance is given by
the equation:
• When resistors are connected in series, the:
-- current through each resistor is the same
1
Rparallel
-- sum of the potential differences is equal to the
potential difference provided to the circuit
=
1
R1
+ R1 + ... + R1
n
2
• Complex circuit analysis may require the
calculation of both equivalent series and
equivalent parallel resistances.
-- equivalent effective resistance is equal to the sum
of the individual resistances, given by the equation
Rseries = R1 + R2 + … + Rn.
• A parallel circuit generally draws more power than
a series circuit using the same resistors.
• Parallel circuits allow individual components to be
switched on and off independently.
• Kirchhoff’s voltage law says that the sum of the
voltages in a closed circuit must be zero, ΣV = 0.
• When resistors are connected in parallel, the:
-- voltage across each resistor is the same
• Kirchhoff’s current law says that the amount of
current flowing into a junction is equal to the
current flowing out (ΣI = 0).
-- current is shared between the resistors
KEY QUESTIONS
1
Two 20 Ω resistors are connected in series with a 6 V
battery. What is the voltage drop across each resistor?
A 0.3 V B 3 V C 6 V D 12 V
2
Consider the series circuit below.
3V
5
Calculate the potential difference across, and the
current through, each resistor in the circuit below.
R4 = 10 Ω
R3 = 10 Ω
12 V
100 Ω
250 Ω
50 Ω
a Calculate the current flowing in the circuit. Give your
answer correct to two significant figures.
b Calculate the potential difference across the 100 Ω
resistor in the circuit.
3
4
396
A 20 Ω resistor and a 10 Ω resistor are connected in
parallel to a 5 V battery. Give your answers correct to
two decimal places.
a Calculate the current drawn from the battery.
b Calculate the current flowing through the 20 Ω
resistor.
c Calculate the current flowing through the 10 Ω
resistor.
A 40 Ω resistor and a 60 Ω resistor are connected in
parallel to a battery, with 300 mA flowing through the
40 Ω resistor.
a Calculate the voltage of the battery.
b Calculate the current flowing through the 60 Ω
resistor.
MODULE 4 | ELECTRICITY AND MAGNETISM
R2 = 15 Ω
R1 = 20 Ω
6
Calculate the equivalent resistance of the combination
of resistors shown below.
3Ω
4Ω
5Ω
6Ω
7
Four 20 Ω light bulbs are connected to a 10 V battery.
What is the total power output of the circuit if the light
bulbs are connected:
a in series? b in parallel?
8
Why are household circuits wired in parallel?
A To reduce the amount of expensive copper wire used.
B To reduce the amount of current drawn by the
household.
C To allow appliances to be switched on and off
independently.
D To reduce the amount of electrical energy used by
the household.
Chapter review
KEY TERMS
ammeter
conventional current
current
effective resistance
electric circuit
electricity
electrical potential energy
electron flow
junction
non-ohmic
ohmic
parallel circuit
potential difference
power
resistance
resistor
series circuit
transfer
transform
volt
voltmeter
REVIEW QUESTIONS
Why is water flowing in a pipe a common analogy for
electric current?
A Water does not conduct electricity.
B Water can leak out of a pipe.
C Water is not compressed or lost as it flows through
a pipe.
D Water and electricity do not mix.
2
Calculate the current that flows when 0.23 C of charge
passes a point in a metal circuit each minute.
3
Compare the meaning of the terms ‘conventional
current’ and ‘electron flow’.
4
Two equal resistors are connected in parallel and
are found to have an equivalent resistance of 68 Ω.
Calculate the resistance of each resistor.
5
6
A battery does 2 joules of work on a charge of 0.5
coulombs to move it from point A to point B. Calculate
the potential difference between the two points A
and B.
Which quantities would you need to measure to
calculate the amount of electrical energy used to heat
water using an electric element?
A potential difference, resistance and current
B time, current and charge
C current, time and potential difference
D potential difference and current
7
How much power does an appliance use if it does
2500 J of work in 30 minutes?
8
A 230 V appliance consumes 2000 W of power. The
appliance is left on for 2 hours. What current flows
through the appliance?
9
A student finds that the current through a wire is 5 A
while a voltage of 2.5 V is applied to it. Calculate the
resistance of the wire.
10 A 60 W incandescent globe draws 0.25 A when
connected to a 240 V power supply. Calculate the
resistance of the globe.
12 When 1.5 V is applied across a particular resistor, the
current through the resistor is 50 mA. What is the
resistance of the resistor?
13 Calculate the resistance of the non-ohmic conductor
with the I–V graph shown in the figure below, at the
following voltages:
a 1.0 V
b 7.0 V
c 12.0 V.
4
3
I (A)
1
2
1
2
4
6
V (V)
8
10
12
14 Explain why even a good conductor such as copper
wire provides some resistance to current.
15 A potential difference of 240 V is used to generate a
current of 5 A to heat water for 3 minutes. Calculate
the energy transferred to the water in that time.
16 A laptop computer draws 500 mA from a 240 V
power point. What is the power consumption of
the computer?
17 A hair dryer is designed to produce 1600 W of power
when connected to a 240 V power supply. What is the
resistance of the hair dryer?
11 A current of 0.25 A flows through an 80 Ω resistor.
Calculate the voltage across it.
CHAPTER 13 | ELECTRIC CIRCUITS
397
CHAPTER REVIEW CONTINUED
18 A 3 V torch with a 0.3 A bulb is switched on for 1 minute.
a How much charge has travelled through the
filament in this time?
b How much energy has been used?
c Where has this energy come from?
19 Two resistors, R1 and R2, are wired in series. Which of
the following gives the equivalent series resistance for
these two resistors?
A Rseries = R1 + R2
B
1
Rseries
=
1
R1
+ R1
2
C Rseries = R1 − R2
D Rseries = R1 × R2
20 An electrical circuit is constructed as shown below.
Use the following information about the circuit
diagram to answer the following questions.
The electric cell provides 3 V.
The total resistance RT of the circuit is 8.5 Ω.
R2 has a resistance of 15 Ω.
The total resistance of resistors R1 and R2 is 5.0 Ω.
b Calculate the total equivalent resistance of the
circuit. Assume that the resistance of the meter and
power source can be ignored.
22 Sketch a circuit diagram showing how four 10 Ω
resistors can be connected using a combination of
series and parallel wiring to have a total equivalent
resistance of 10 Ω.
23 A 9 V battery is connected to a voltage divider
consisting of two resistors with values of 600 Ω and
1200 Ω. What would the voltage across the 600 Ω
resistor be?
24 A voltage divider is constructed using a 20 V battery,
a 400 Ω resistor and a variable resistor, R, as shown
in the diagram. What should the resistance of R be in
order to produce Vout = 8 V?
+
R
20 V
Vout
R1
R2
400 Ω
R3
a Find the value of R3.
b Find the current through R3.
c Find the potential difference across the parallel pair
R1 and R2.
d Find the current through R2.
e Find the current through R1.
f Find the value of R1.
21 Eight equal-value resistors are connected between
points P and Q. The value of each of these resistors
is 20.0 Ω.
P
Q
a The circle in the circuit diagram represents either
an ammeter or a voltmeter. Identify which type of
meter this should be.
398
MODULE 4 | ELECTRICITY AND MAGNETISM
A
B
C
D
400 Ω
600 Ω
800 Ω
1200 Ω
25 A circuit consists of a 12 V battery and three 20 Ω light
bulbs. The bulbs are initially connected in series.
a Calculate the power output of the circuit.
b The circuit is changed so that the bulbs are
connected in parallel. Calculate the power output of
the circuit.
c Compare the power drawn in the parallel circuit
with that of the series circuit.
26 After completing the activity on page 384, reflect
on the inquiry question: How do the processes of
the transfer and the transformation of energy occur
in electric circuits? In your response, compare the
water circuit analogy to electric circuits. Sketch two
electric circuits that would work the same as the two
water circuits analysed.
CHAPTER
Magnetism
While naturally occurring magnets had been known for many centuries, by the
early 19th century there was still no scientifically proven way of creating an artificial
magnet. In 1820, the Danish physicist Hans Christian Ørsted developed a scientific
explanation for the magnetic effect created by an electric current.
In this chapter you will investigate the concept of magnetism, how materials become
magnetised, how to describe magnetic fields through the use of field diagrams and
how to calculate magnetic fields in relation to electric current.
Content
INQUIRY QUESTION
How do magnetised and magnetic objects interact?
By the end of this chapter you will be able to:
• investigate and describe qualitatively the force produced between magnetised and
magnetic materials in the context of ferromagnetic materials (ACSPH079)
• use magnetic field lines to model qualitatively the direction and strength of
magnetic fields produced by magnets, current-carrying wires and solenoids, and
relate these fields to their effect on magnetic materials that are placed within
them (ACSPH083) ICT
• conduct investigations into and describe quantitatively the magnetic fields
produced by wires and solenoids, including: (ACSPH106, ACSPH107)
-- B =
µ0I
2πr
-- B =
µ0NI
L
ICT
ICT
N
N
• investigate and explain the process by which ferromagnetic materials become
magnetised (ACSPH083)
• apply models to represent qualitatively and describe quantitatively the features of
magnetic fields. ICT N
Physics Stage 6 Syllabus © NSW Education Standards Authority
for and on behalf of the Crown in right of the State of NSW, 2017.
14.1 Magnetic materials
Try breaking a magnet in half. All you get is two smaller magnets, each with its own
north and south poles. No matter how many times you break the magnet and how
small the pieces are, each will be a separate magnet with two poles. Because magnets
always have two poles, they are said to be dipolar.
Magnets are dipolar, and a magnetic field is a dipole field (Figure 14.1.1). This
is similar to electric charges, where a positive and negative charge in close proximity
to each other are said to form a dipole. Unlike electrostatics, though, you cannot
have a single magnetic pole.
N
FIGURE 14.1.1
S
N
S
N
S
N
S
N
S
N
S
N
S
Magnets are always dipolar.
MAGNETISM
Magnetism is a physical phenomenon caused by magnets that results in a field that
attracts or repels other magnetic materials.
Every material experiences magnetism to some extent, some more strongly than
others. The magnetic effect most people are familiar with is the attraction of iron or
other magnetic materials to a magnet, as seen in Figure 14.1.2.
geographic
North Pole
magnetic
North Pole
S
FIGURE 14.1.2
N
magnetic
South Pole
geographic
South Pole
FIGURE 14.1.3 The Earth acts somewhat like
a huge bar magnet. The south pole of this
imaginary magnet is near the geographic North
Pole and is the point to which the north pole of
a compass appears to point.
400
A bar magnet attracting drawing pins.
A suspended magnet that is free to move will always orientate itself in a north–
south direction.That’s basically what the needle of a compass is—a freely suspended,
small magnet. If allowed to swing vertically as well, then the magnet will tend to tilt
vertically. The vertical direction (upwards/downwards) and the magnitude of the
angle depend upon the distance of the magnet from either of the Earth’s poles.
As you can see in Figure 14.1.3, the Earth itself has a giant magnetic field
around it.
The names for the poles of a magnet derive from early observations of magnets
orientating themselves with the Earth’s geographic poles.
Initially, the end of the magnet pointing toward the Earth’s geographic north
was denoted the North Pole, and compasses are thus marked with this end as
north. However, it is now known that the geographic North Pole and the magnetic
North Pole are slightly apart in distance, and the same applies to the geographic
South Pole and the magnetic South Pole.
MODULE 4 | ELECTRICITY AND MAGNETISM
PHYSICS FILE ICT
‘Flipping’ poles
The Earth’s magnetic poles are not static like their geographic counterparts. For
many years, the magnetic North Pole had been measured as moving at around 9 km
per year (Figure 14.1.4). In recent years that has accelerated to an average of 52 km
per year. Once every few hundred thousand years the magnetic poles actually flip in
a phenomenon called ‘geomagnetic reversal’, so that a compass would point south
instead of north. The Earth is well overdue for the next flip, and recent measurements
have shown that the Earth’s magnetic field is starting to weaken faster than in the
past, so the magnetic poles may be getting close to a ‘flip’. While past studies have
suggested such a flip is not instantaneous—it would take many hundreds if not a few
thousands of years—some more recent studies have suggested that it could happen
over a significantly shorter time period.
1900
1996
outer core
FIGURE 14.1.4 Diagram of Earth’s interior and the movement of magnetic north from 1900 to
1996. The Earth’s outer core is believed to be the source of the geomagnetic field.
FERROMAGNETIC MATERIALS
Some materials experience magnetism more than other materials. For example, iron,
cobalt and nickel experience magnetism more than aluminium. These materials are
known as ferromagnetic materials.
Ferromagnetic materials are used in many common applications where magnetic
materials are required. Examples include the hard disk drive of a computer
(Figure 14.1.5) and the magnetic stripe on credit cards and EFTPOS cards.
Ferromagnetic materials are used in many industrial applications as well, such as
motors, generators and in electric power transmission and distribution.
FIGURE 14.1.5 Ferromagnetic materials are
used in many common applications, such as
computer hard disk drives.
CHAPTER 14 | MAGNETISM
401
HOW DO MATERIALS BECOME MAGNETISED?
(a)
(b)
FIGURE 14.1.6 (a)
The magnetic fields in
separate magnetic domains point in different
directions. As a result, their magnetic fields
cancel out, resulting in an un-magnetised
material. (b) The magnetic fields in separate
magnetic domains point in the same direction
(align), resulting in a magnetised material.
Like magnetic poles repel each
other; unlike magnetic poles
attract each other.
The bulk piece of a ferromagnetic material is divided into magnetic domains. A
magnetic domain is a region in the material where the magnetic field is aligned.
Most often, ferromagnetic materials are found in an un-magnetised state. This
is because before a ferromagnetic material becomes magnetised, the magnetic fields
in separate magnetic domains point in different directions (Figure 14.1.6a). As a
result, their magnetic fields cancel each other out.
By applying an external magnetic field to the un-magnetised material,
the individual domains align with the external magnetic field. Aligning all the
magnetic domains in a ferromagnetic material would create a larger magnetic field
(Figure 14.1.6b) resulting in a strongly magnetised material. When the external
magnetic field is removed, the individual magnetic domains remain fixed in their
new orientation, and the newly aligned domains produce a uniform resultant
magnetic field. This is how a ferromagnetic material is magnetised.
Forces between magnetic materials
A magnetic material produces a non-contact force (or force mediated by a field)
around itself. The magnetic material creates the magnetic field, which in turn
produces a magnetic force. This force is what causes magnetic materials to stick
together, such as magnets on your refrigerator.
If you experiment with a magnet yourself, you will find that each end of a
magnet behaves differently, particularly when interacting with another magnet.
One end is attracted while the other is repelled (Figure 14.1.7). Each end of a
magnet is referred to as a magnetic pole. Like magnetic poles, such as two north
poles, experience a repulsive force, whereas unlike poles (north and south poles)
experience an attractive force.
FIGURE 14.1.7
stick together.
402
MODULE 4 | ELECTRICITY AND MAGNETISM
Opposite poles in a magnet produce an attractive force, causing the two magnets to
14.1 Review
SUMMARY
• Like magnetic poles repel, and unlike magnetic
poles attract.
• The magnetic field in separate magnetic domains
points in different directions in an un-magnetised
material.
• Magnetic poles exist only as dipoles, having both
north and south poles. A single magnetic pole
(monopole) is not known to exist.
• The magnetic field in separate magnetic domains
points in the same direction in a magnetised
material.
• Ferromagnetic materials, such as iron, cobalt
and nickel, are materials that can easily become
magnetised.
KEY QUESTIONS
1
Repeatedly cutting a magnet in half always produces
magnets with two opposite poles. From this
information, which of the following can be deduced in
relation to the poles of a magnet?
A Magnets are easily sliced in half.
B All magnets are dipolar.
C When the magnets are cut, the poles are split in half.
D All split magnets are monopolar.
2
State three ferromagnetic materials.
3
Describe the forces that occur between like and
opposite poles in a pair of magnets.
4
The bulk piece of a ferromagnetic material is divided
into magnetic domains. If the majority of the magnetic
domains have magnetic fields that align, would that
increase or decrease the magnetic field strength of the
material?
5
Which definition best describes a magnetic domain?
A A region in the material where the magnetic field
does not align.
B A region in the material where the magnetic field
aligns.
C Multiple regions in the material where the magnetic
fields do not align.
D Multiple regions in the material where the magnetic
fields align.
6
By applying an external magnetic field to an unmagnetised material, the individual domains align
with the external magnetic field. What happens to the
material when the external magnetic field is removed?
7
A magnetic material creates a magnetic field, which
in turn produces a magnetic force. This force is what
causes magnetic materials to stick together, such as
magnets on your refrigerator. Is this force a contact or
non-contact force?
CHAPTER 14 | MAGNETISM
403
PHYSICS INQUIRY
N CCT
Magnetic doorbell
How do magnetised
and magnetic objects
interact?
COLLECT THIS…
•
straw with large diameter
•
metal rod (that can fit inside
the straw)
•
long electrical wire
•
power pack or 9 V battery and
switch
•
bike bell (or similar)
14.2 Magnetic fields
In Chapter 12 you saw that point charges and charged objects produce an electric
field in the space that surrounds them. For this reason, charged bodies within the
field experience a force. The direction of the electric field is determined by the
direction of that force.
Magnets also create fields. If you do a simple test like placing a pin near a magnet,
you will observe that the pin is pulled toward the magnet. This shows that the space
around the magnet must therefore be affected by the magnet.
If you sprinkle iron filings on a piece of clear acetate that is held over a magnet,
you will observe that the magnetic field will be clearly defined (Figure 14.2.1). The
iron filings line up with the field, showing clear field lines running from one end of
the magnet to the other.
(a)
(b)
DO THIS…
1 Place the metal rod in the
straw.
2 Wrap the wire around the straw
(tight enough for it to stay in
place but not so tight that it
would restrict the motion of the
metal rod).
FIGURE 14.2.1 Iron filings sprinkled around magnets (a) with unlike poles close together and (b) with
like poles close together. The patterns in the fields show the attraction and repulsion between poles,
respectively.
3 Connect the ends of the wire to
the power supply, keeping the
electricity switched off initially.
VECTOR FIELD MODEL FOR MAGNETIC FIELDS
4 Position the metal rod towards
one end of the straw, and place
the bell at the other end.
The diagram in Figure 14.2.2 shows the magnetic field associated with a simple bar
magnet. The magnetic field around the bar magnet can be defined in vector terms,
specifying both direction and magnitude.
5 Turn on the electricity for a
short time and observe the
system. Do not leave the
electricity on as the wires will
heat up.
N
6 Change the position of the rod
and bell and repeat step 5.
RECORD THIS…
S
Describe the magnetic field that is
produced by the coil of wire.
Present a diagram of your doorbell
design.
REFLECT ON THIS…
How do magnetised and magnetic
objects interact?
What variables affect the doorbell?
404
FIGURE 14.2.2 The field lines around and inside a bar magnet. The lines show the direction of the
force on an (imaginary) single north pole.
The direction of the magnetic field at any point is the direction that a compass
would point if placed at that point—that is, towards the magnetic South Pole. This
is also the direction of the force the magnetic field would exert on an (imaginary)
single north pole.
MODULE 4 | ELECTRICITY AND MAGNETISM
Denser (closer) lines indicate a stronger magnetic field. As the distance from the
magnet increases, the magnetic field is spread over a greater area and its strength at
any point decreases. This is the same as with electric fields, as seen in Chapter 12.
The strength and direction associated with the magnetic field at any point signifies
that it is a vector quantity.
DS
4.1
GO TO ➤ Section 12.2, page 345
The magnetic field (B) is a vector quantity as it has both a magnitude and
a direction.
The strength, or vector magnitude, of the magnetic field is in units of tesla (T) and
is denoted by the variable B.
The fields between magnets are dependent on whether like or unlike poles
are close together, the distance the poles are apart and the relative strength of the
magnetic field of each magnet. Iron filings or small plotting compasses can be used
to visualise the field between and around the magnets, as shown in Figure 14.2.3.
N
S
FIGURE 14.2.3 Plotting field lines around a bar magnet. Small plotting compasses are placed around
the magnet. Field lines are drawn linking the direction each compass points in, creating field lines
that run from the north pole to the south pole of the magnet.
Because the Earth has a giant magnetic field around it, you can also predict how
compasses will orient themselves around the Earth—they will orient themselves
along the magnetic field lines. In Figure 14.2.3, note the direction of the magnetic
field close to either pole, where the magnetic field lines run almost vertically. Magnets
placed near the Earth’s magnetic poles will behave in the same way.
Different-shaped magnets produce different-shaped fields. The diagram in
Figure 14.2.4 shows the magnetic field plotted for a horseshoe magnet.
N
S
FIGURE 14.2.4 The horseshoe magnet has two unlike poles close to each other. This creates a very
strong magnetic field.
CHAPTER 14 | MAGNETISM
405
The resultant direction of the magnetic field at a particular point is the vector
addition of each individual magnetic field acting at that point.
Figure 14.2.5a shows a bar magnet and its associated magnetic field lines. When
two magnets are placed close together, two situations may arise. If the poles are
unlike as in Figure 14.2.5b, then attraction will occur between them and a magnetic
field will be created that extends between the two poles. On the other hand, if like
poles are near each other as in Figure 14.2.5c, they will repel each other. In this
situation, there is a neutral point (no magnetic field) between the two poles.
(a)
(b)
(c)
N
S
S
S
N
N
S
N
neutral or
null point
N
S
FIGURE 14.2.5 Magnetic field lines plotted for (a) a bar magnet, (b) opposite poles of bar magnets in
close proximity and (c) like poles of bar magnets in close proximity.
As the bar magnets in Figure 14.2.5 have a fixed strength and position, the
associated magnetic fields are static. Varying the magnetic field strength by changing
the magnets or varying the relative position of the magnets produces a changing
magnetic field.
WS
4.8
MAGNETIC FIELDS AND CURRENT-CARRYING WIRES
In 1820, Hans Christian Ørsted discovered that an electric current could produce a
magnetic field. His work established the initial ideas behind electromagnetism. Since
then, our understanding and application of electromagnetism has developed to the
extent that much of our modern way of living relies upon it.
Ørsted found that when he switched on the current from a voltaic pile
(a simple early battery), a nearby magnetic compass needle moved. Intrigued by
this observation, he carried out further experiments that demonstrated that it was
the current from the voltaic pile that was affecting the compass movement. His
experiments showed that the stronger the current, and the closer the compass was
to it, the greater the observed effect was. These observations led him to conclude
that the electric current was creating a magnetic field. It is believed that the Earth’s
magnetic field is created by a similar effect—circulating electric currents in the
Earth’s molten metallic core.
A circular magnetic field is created around a current-carrying wire.
The magnetic field around a current-carrying wire can be seen in Figure 14.2.6a.
A compass aligns itself at a tangent to the concentric circles around the wire (i.e.
the magnetic field). This is seen in Figure 14.2.6b. The stronger the current and the
closer the compass is to the wire, the greater the effect.
406
MODULE 4 | ELECTRICITY AND MAGNETISM
(a)
(b) Direction of
current
SKILLBUILDER
N
N
N
N
(a) The magnetic field around a current-carrying wire. The iron filings align with the
field to show the circular nature of the magnetic field. (b) Small compasses indicating the direction
of the field.
FIGURE 14.2.6
The magnetic field is perpendicular to the current-carrying wire and the
direction of the field depends on the current direction. There’s a simple and easy
way to determine the direction of the magnetic field, which is commonly referred
to as the right-hand grip rule.
PHYSICSFILE
Right-hand grip
rule
Imagine that you have grasped the
conducting wire with your right
hand with your thumb pointing in
the direction of the conventional
electric current, I (positive to
negative). Curl your fingers around
the wire. The magnetic field will
be perpendicular to the wire and
in the direction your fingers are
pointing, as shown in Figure 14.2.7.
S
Sharks and magnetism
Sharks have special sensory organs called electroreceptors. These electroreceptors
help sharks sense electric fields in the water. When an electrically conducting object
moves within a magnetic field (e.g. Earth’s magnetic field), an electric field is induced.
In addition, small electric fields are produced by muscle contractions in living creatures.
Sharks may be able to detect prey in the water, as the prey swim through the Earth’s
magnetic fields creating electric fields.
N
I
S
FIGURE 14.2.7 The right-hand grip rule
can be used to find the direction of
the magnetic field around a currentcarrying wire, when the direction of the
conventional current, I, is known.
Remember from Chapter 13 that
conventional current, I, is in the
opposite direction to electron flow.
FIGURE 14.2.8
Sharks are able to hunt their prey using electroreceptors.
CHAPTER 14 | MAGNETISM
407
Worked example 14.2.1
DIRECTION OF THE MAGNETIC FIELD
A current-carrying wire runs horizontally across a table. The conventional current
direction, I, is running from left to right.
What is the direction of the magnetic field created by the current?
Thinking
Working
Recall that the right-hand grip
rule indicates the direction of the
magnetic field.
Hold your hand with your fingers
aligned as if gripping the wire.
Point your thumb to the right in the
direction of the current flow.
fingers point down on
the front side of the wire
current
Describe the direction of the field
in relation to the reference object
or wire, in a way that can be readily
understood by a reasonable reader.
The magnetic field direction is
perpendicular to the wire and runs up
from the back of the wire, over the top
towards the front of the wire.
Worked example: Try yourself 14.2.1
DIRECTION OF THE MAGNETIC FIELD
A current-carrying wire runs along the length of a table. The conventional current
direction, I, is running toward an observer standing at the near end.
What is the direction of the magnetic field created by the current as seen by
the observer?
Magnetic fields between parallel wires
(a)
I
S
N
I
(b)
S
S
I
S
N
N
N
I
FIGURE 14.2.9 (a) Two current-carrying wires
attract when current runs through them in the
same direction. This is because the magnetic
fields between the wires are in opposite
directions. (b) Two current-carrying wires
repel when the current passes through them
in opposite directions. This is because the
magnetic fields are in the same direction.
408
Two current-carrying wires arranged parallel to each other each have their own
magnetic field. The direction of the magnetic field around each wire is given by the
right-hand grip rule. If the two wires are brought closer together, their associated
magnetic fields will interact, just as any two regular magnets would interact. The
interaction could result in either an attraction or repulsion of the wires, depending
on the direction of the magnetic fields between them (Figure 14.2.9). When the
magnetic fields are in the opposite directions, this represents unlike poles, and so
the wires attract. When the magnetic fields are in the same direction, the wires repel.
When a field is running directly into or out of the plane of the page, dots are used
to show a field coming out of the page and crosses are used to depict a field running
directly into the page. This convention was adopted from the idea of viewing an
arrow. The dot is the point of the arrow coming toward you, and the cross represents
the tail feathers of the arrow as it travels away. Figure 14.2.10 shows the same
current-carrying wires from 14.2.9a using the dot and arrow representation.
3D FIELDS
Field lines can also be drawn for more-complex, 3D fields such as that around
the Earth or those around current-carrying loops and coils. Even in more-complex
fields, the right-hand grip rule is still applicable, as you can see in Figures 14.2.11
and 14.2.12.
MODULE 4 | ELECTRICITY AND MAGNETISM
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
I
FIGURE 14.2.10
×
×
×
×
×
×
×
×
I
A 2D representation of the field around a current-carrying wire.
(b)
(a)
I
A 3D representation of the
magnetic field around a loop of wire in the
plane of the page. The blue arrows show the
direction of the magnetic field. Notice that the
magnetic field has a circular shape, with no field
lines crossing.
FIGURE 14.2.12
I
I
FIGURE 14.2.11 The magnetic field lines around (a) a single current loop and (b) a series of loops.
The blue arrows indicate the direction of the magnetic field. The more concentrated the lines are
inside the loops, the stronger the magnetic field is in this region.
The direction of a magnetic field can be shown with a simple arrow on a field line
when the field is travelling within the plane of the page (as shown in Figure 14.2.11).
Figure 14.2.12 shows a 3D representation of the magnetic field around a loop of
wire. This same loop can also be represented in two dimensions using the dot-andarrow convention (Figure 14.2.13).
The strength of a field is depicted by varying the density of the lines or dots and
crosses. Showing lines coming closer together indicates a strengthening of a field, and
lines farther apart indicates a weaker field. More densely placed dots or crosses also
show a stronger area of the field. This is referred to as a non-uniform magnetic field.
As the magnetic fields associated with current-carrying coils depend on the size
of the current, the associated magnetic field may also change over time, either in
magnitude, or, if the current is reversed, in direction.
THE MAGNETIC FIELD AROUND A SOLENOID
x x x x x
x
x x x
x x
x x
x
x
x
x
x x
x x
x x x x
x x x x
x x
x
x
x
x x
x
x
x
x
x x x
x
x x x
x
x x x
x x x
x
FIGURE 14.2.13 A 2D representation of
the same current-carrying loop depicted in
Figure 14.2.12. Areas where the magnetic field
is stronger are shown with a greater density of
dots and crosses.
If many loops are placed side by side, their fields all add together and there is a
much stronger effect. This can easily be achieved by winding many turns of wire
into a coil termed a solenoid. The field around the solenoid is like the field around
a normal bar magnet. The direction of the overall magnetic field can be determined
by considering the field around each loop and, in turn, the field around the currentcarrying wire making up the loop. The direction of the field of the solenoid depends
on the direction of the current in the wire making up the solenoid.
This is explained in Figure 14.2.14.
FIGURE 14.2.14 This solenoid has an effective ‘north’ end at the left and a ‘south’ end at the right. A
compass would point in the direction of the field lines.
CHAPTER 14 | MAGNETISM
409
CREATING AN ELECTROMAGNET
FIGURE 14.2.16 A large electromagnet being
used to lift waste iron and steel at a scrapyard.
Valuable metals such as these are separated and
then recycled.
The earliest magnets were all naturally occurring. If you wanted a magnet, you
needed to find one. They were regarded largely as curiosities. Hans Christian
Ørsted’s discoveries made it possible to manufacture magnets, making the widescale use of magnets possible.
An electromagnet, as the name infers, runs on electricity. It works because an
electric current produces a magnetic field around a current-carrying wire. If the
conductor is looped into a series of coils to make a solenoid, then the magnetic field
is concentrated within the coils. The more coils, the stronger the magnetic field and,
therefore, the stronger the electromagnet.
The magnetic field can be strengthened further by wrapping the coils around a
core. Normally, the atoms in materials like iron point in random directions and the
individual magnetic fields tend to cancel each other out. However, the magnetic field
produced by coils wrapped around an iron core forces the atoms within the core
to point in one direction. Their individual magnetic fields add together, creating
a stronger magnetic field. This is similar to the concept of making a permanent
magnet, discussed in Section 14.1.
The strength of an electromagnet can also be changed by varying the amount of
electric current that flows through it.
The direction of the current creates poles in the electromagnet. The poles of
an electromagnet can be reversed by reversing the direction of the electric current.
Today, electromagnets are used directly to lift heavy objects (Figure 14.2.16), as
switches and relays, and as a way of creating new permanent magnets by aligning
the atoms within magnetic materials.
+ ADDITIONAL
Transformers and electromagnets
Electromagnetism, ferromagnetic materials and magnetic
fields from current-carrying wires are very common and
have many practical uses. One common application is
a power transformer. You will often see these on the
power lines near your house (Figure 14.2.15). This will be
covered in more detail in Physics 12.
FIGURE 14.2.15 Electric power line poles carry the current and power
to houses, hold transformers (grey boxes), and sometimes carry
telephone cables.
410
MODULE 4 | ELECTRICITY AND MAGNETISM
Transformers use electromagnetic principles to convert
high voltages to lower voltages safe for household use.
When electric power is generated at a power plant, it is
transmitted at very high voltages (hundreds of thousands
of volts), to prevent power loss over long distances. For the
same transmitted power, increasing the voltage decreases
the current (recall Ohm’s law from Chapter 13 V = IR).
Decreasing the current in the transmitted power lines
reduces the power loss, as the power loss is P = VI = I2R,
where I is the current and R is the resistance of the power
line. Depending upon the end user’s requirements, the
transmitted high voltage needs to be lowered. For household
appliances and circuits, 240 V is the requirement.
Inside a transformer is a ferromagnetic core that has two
windings: a primary and a secondary. The high-voltage
wires are wound around the primary side. The low-voltage
wires are wound around the secondary side. The current
in the wire from the high-voltage side creates a magnetic
field, much like the example shown with the solenoid.
That magnetic field creates a current in the wires on the
secondary side, which then feed the household circuits. The
number of windings on the primary and secondary sides
determine whether the voltage is increased or decreased.
14.2 Review
SUMMARY
• The direction of the magnetic field is from the
magnetic north pole to the magnetic south pole.
• An electrical current produces a magnetic
field that is circular around a current-carrying
conductor. The direction of the field is given by
using the right-hand grip rule when considering
the direction of the conventional current.
• Denser (closer) lines indicate a relatively stronger
magnetic field.
• A magnetic field associated with a constant
magnetic field is static. Where the magnetic field
is changing, such as that associated with an
alternating current direction, the magnetic field
will also be changing.
• More complex fields can be determined by
applying the right-hand grip rule to the loops or
coils making up the current-carrying conductor in
a solenoid.
• The strength of a magnet can be increased by
wrapping a current-carrying wire around the
magnet to create an electromagnet.
• A uniform distribution of field lines represents a
uniform magnetic field. A non-uniform field, such
as that around a non-circular coil, is shown by
variations in the separation of the field lines.
KEY QUESTIONS
1
The field around a particular current-carrying loop
shows a variation in the strength of its magnetic field,
as depicted below. The current in the loop itself is
being switched on and off but is constant in direction
and size.
x x x x x
x
x
x x x
x x
x x
x x x x x x
x
x x
x x x
x x x x
x x
x
x
x
x
x x
x
x
x
x x x
x
x
x x
x
x
x x
x x x
Which of the following best describes the resulting
magnetic field of the loop?
A a reversed magnetic field
B a static magnetic field
C a non-uniform magnetic field
D a uniform magnetic field
2
A current-carrying wire runs horizontally across a table.
The conventional current direction, I, is running from
right to left. Draw a diagram showing the direction of
the magnetic field around the wire.
3
The following diagram shows a current-carrying
solenoid.
B
A
I
I
Which end (A or B) represents the north pole of
this solenoid?
CHAPTER 14 | MAGNETISM
411
4
Two strong bar magnets with magnetic fields of equal
strength are arranged as shown.
N
N
A
B C
1.0 cm
1.5 cm
1.5 cm
1.0 cm
S
N
S
E
W
S
Ignoring the magnetic field of the Earth, what is the
approximate direction of the resulting magnetic field:
a at point A?
b at point C?
c at point B?
5
The figure below shows a cross-sectional view of a
long, straight, current-carrying conductor, with its axis
perpendicular to the plane of the page. The conductor
carries an electric current into the page.
A
D
N
B
W
C
412
MODULE 4 | ELECTRICITY AND MAGNETISM
E
S
a What is the direction of the magnetic field produced
by this conductor at each of the points A, B, C and D?
b The direction of the current in the conductor is now
changed so that it is carried out of the page. What is
the direction of the magnetic field produced by this
conductor at the four points A, B, C and D?
14.3 Calculating magnetic fields
In the previous section, we saw that a current-carrying wire creates a magnetic field.
In this section, we will determine how to calculate the magnitude of the magnetic field.
AMPERE’S LAW
Ampere’s law states that the sum of all the magnetic field elements, B, that make
up the circle surrounding the wire is equal to the product of the current in the wire
and the permeability of free space (a constant µ0). Ampere’s law can be used to
calculate the magnetic field surrounding a current-carrying wire, as shown in the
following equation:
µ I
The permeability of free space,
µ0, is approximately equal to
1.257 × 10−6 N A−2 (newtons per
ampere squared).
If you equate the units in the
equation for Ampere’s law using
dimensional analysis, you find
that 1 tesla (T) is equivalent to
1 N A−2 m−1.
B = 2π0 r
where:
B is magnetic field strength (in T)
µ0 is the permeability of free space (in N A−2)
I is the current (in A)
r is the distance from the wire (in m).
PHYSICSFILE ICT
Loudspeakers
Ampere’s law can be further expanded to calculate the magnetic field surrounding
a solenoid. The magnetic field at a point well inside the solenoid is uniform and
independent of the length or diameter of the solenoid. Instead, it is dependent upon
the number of turns of the coil (N) per unit length (L) of the solenoid. This is shown
in the following equation:
B=
µ0NI
L
where N is the number of turns per unit length (L).
The magnetic field strength is independent of the total length of the solenoid.
This means that increasing the number of turns of a solenoid can be used to
increase its magnetic field strength (i.e. B ∝ N). It is important to keep in mind that
this is the number of turns per unit length. So two solenoids with the same current
but different lengths can still have the same magnetic field strength. For example,
solenoid A may have 100 turns in one metre, while solenoid B has 50 turns in 0.5 m,
but with the same current these will produce the same magnetic field strength.
Solenoids can be used to generate nearly uniform magnetic fields similar to the
fields produced by a bar magnet. There are many practical applications of solenoids,
such as in electromagnets, power generation and distribution, and electric motors.
These topics will be covered in much greater detail in Year 12 Physics.
Inside a loudspeaker (Figure 14.3.1)
is a permanent magnet and a coil
wrapped around a metal rod.
When the loudspeaker is connected
to a music source (stereo, iPod etc.),
electric signals from the source
run through the coil inside the
loudspeaker. The electric signals are
sent in the form of electric current,
which turns the coil wrapped around
the metal rod into an electromagnet.
As the electric current flows back
and forth in the coil, the magnetic
field in the electromagnet either
attracts or repels the permanent
magnet. This causes the metal rod
to move backwards and forwards,
pushing or pulling the cone, causing
sound pressure to change, and you
(the listener) to hear the sound.
permanent
moving magnet
coil
Worked example 14.3.1
MAGNITUDE OF THE MAGNETIC FIELD
An electric current of 10 A is passed through a wire.
What is the magnetic field created by this current at a distance of 4 cm from
the wire?
Thinking
Working
Recall the formula used to calculate the magnetic
field surrounding a current-carrying wire.
B=
Substitute known values into the equation.
× 10 × 10
B = 1.257
2 π × 0.04
Solve for the magnetic field B.
B = 5 × 10−5 T
cone
current
from
amplifier
µ0I
2πr
−6
FIGURE 14.3.1 A diagram of a loudspeaker
with an electromagnet inside.
CHAPTER 14 | MAGNETISM
413
Worked example: Try yourself 14.3.1
MAGNITUDE OF THE MAGNETIC FIELD
A charged wire carries a current of 2 A.
What is the magnetic field created by the wire at a distance of 10 cm?
Worked example 14.3.2
MAGNITUDE AND DIRECTION OF THE MAGNETIC FIELD
A current-carrying wire runs horizontally across a table. The conventional current
direction, I, is running from right to left. The current-carrying wire produces a
magnetic field of 6 × 10−6 T measured 5 cm from the wire.
a What current must the wire be carrying to produce that field?
Thinking
Recall the formula used to calculate
the magnetic field surrounding a
current-carrying wire.
Working
B=
µ0I
2π r
Rearrange the formula to solve for the
unknown quantity.
I=
2 π rB
µ0
Substitute known values into the
equation.
I=
Solve for the current I.
I = 1.5 A
2 π × 0.05 × 6 × 10−6
1.257 × 10−6
b Draw a diagram showing the direction of the magnetic field around the wire.
Thinking
Working
Use the right-hand rule to determine
the direction of the magnetic field.
The direction of the magnetic field
created by the current is perpendicular
to the wire and runs up the front of the
wire then down the back when looking
from the front of the wire.
Draw a diagram of the magnetic field
lines.
I
S
N
Worked example: Try yourself 14.3.2
MAGNITUDE AND DIRECTION OF THE MAGNETIC FIELD
A current-carrying wire runs horizontally across a table. The conventional current
direction, I, is running from left to right. The current-carrying wire produces a
magnetic field of 3 × 10−6 T measured 15 cm from the wire.
a What current must the wire be carrying to produce that field?
b Draw a diagram showing the direction of the magnetic field around the wire.
414
MODULE 4 | ELECTRICITY AND MAGNETISM
Worked example 14.3.3
MAGNITUDE OF THE MAGNETIC FIELD IN A SOLENOID
A solenoid of length 1 m with N = 10 turns carries a current of 10 A.
a What is the magnetic field created by this current at a point well inside
the solenoid?
Thinking
Working
Recall the formula used to calculate
the magnetic field in a solenoid.
B=
Determine what values to use, in
SI units.
µ0NI
L
µ0 = 1.257 × 10−6 T A−1
I = 10 A
N = 10
L = 1m
−6
Substitute in known values into the
equation.
B = 1.257 × 101
× 10 × 10
Solve for the magnetic field B.
B = 1.3 × 10−4 T
b If the number of turns is increased to 20, what effect would that have on the
magnetic field strength?
Thinking
Working
Recall the formula for calculating the
magnetic field in a solenoid.
Using the formula to calculate the
magnetic field in a solenoid, determine
the effect of doubling N.
B=
µ0NI
L
B=
µ0NI
L
N2 = 20 = 2N1, then:
B2 =
µ0 × 2N1I
L
= 2 × B1
The magnetic field would double.
Worked example: Try yourself 14.3.3
MAGNITUDE OF THE MAGNETIC FIELD IN A SOLENOID
A solenoid of length 1 m with N = 100 turns carries a current of 2.0 A.
a What is the magnetic field created by this current at a point well inside the
solenoid?
b If the number of turns is decreased to 50, what effect would that have on
the magnetic field strength?
WS
4.9
PA
4.5
PA
4.5
CHAPTER 14 | MAGNETISM
415
14.3 Review
SUMMARY
• A current-carrying wire produces a magnetic field
in the form of concentric circles surrounding the
wire. The magnitude of that magnetic field is
defined by Ampere’s law.
• The magnetic field at a point well inside a solenoid
is uniform and independent of the length or
diameter of the solenoid. Instead, it is dependent
upon the number of turns of the coil (N) per unit
length (L) of the solenoid.
• Ampere’s law states that the sum of all the
magnetic field elements that make up the
circle surrounding the wire is the product of
the current in the wire (I) and the permeability
of free space (µ0).
B=
B=
µ0NI
L
• The more tightly wound the wire around a solenoid,
the greater the magnetic field.
• Solenoids can be used to generate nearly uniform
magnetic fields, similar to the fields generated by
a bar magnet.
µ0I
2π r
KEY QUESTIONS
1
A current-carrying wire produces a magnetic field in
the form of concentric circles surrounding the wire.
Is the magnetic field strength directly or inversely
proportional to the magnitude of the current?
2
True or false: The magnetic field at a point well inside
a solenoid is dependent on the length or diameter of
the solenoid.
3
Assume the magnetic field produced by a solenoid is
6 × 10−6 T.
a If you increased the number of turns per unit length
of the solenoid, would the magnetic field strength
increase or decrease?
b If the number of turns per unit length of the
solenoid is increased by a factor of four, what would
the resulting magnetic field strength be?
c If the number of turns per unit length of the
solenoid is decreased by a factor of two, what would
the resulting magnetic field strength be?
4
An electromagnet is an example of a solenoid. Assume
you wish to increase the strength of the electromagnet
by increasing the magnetic field strength in the
solenoid. What two variables could you change to
increase the magnetic field strength?
416
MODULE 4 | ELECTRICITY AND MAGNETISM
5
A current-carrying wire carries a current of 1 A. What is
the magnetic field created by this current at a distance
of 2 cm from the wire?
6
A current-carrying wire produces a magnetic field of
2 × 10−6 T measured 10 cm from the wire. What current
must the wire be carrying to produce that field?
7
A solenoid carrying a current of 10 A produces a
magnetic field of 6 × 10−4 T.
a If the current was doubled to 20 A and the number
of turns per unit length remained the same, what
would the resulting magnetic field strength be?
b If the current was halved to 5 A and the number
of turns per unit length remained the same, what
would the resulting magnetic field strength be?
c If the current is doubled to 20 A and the number
of turns per unit length doubled, what would the
resulting magnetic field strength be?
d If the current is halved to 5 A and the number of
turns per unit length were halved, what would the
resulting magnetic field strength be?
8
Use dimensional analysis to confirm that the unit
of magnetic field strength—Tesla—is equivalent to
kg A−1 s−2.
Chapter review
KEY TERMS
Ampere’s law
dipole
electromagnet
ferromagnetic material
magnetic
magnetic domain
magnetic field
magnetism
pole
right-hand grip rule
solenoid
voltaic pile
REVIEW QUESTIONS
1
Choose the correct statements from the options below,
relating to magnetic poles.
A opposite poles attract
B like poles attract
C opposite poles repel
D like poles repel
2
Define the term ‘magnetic dipole’.
3
Explain why ferromagnetic materials make stronger
permanent magnets.
4
What is the most appropriate material to use for the
core of an electromagnet?
5
What distinguishes the force between two magnetic
objects from the force between two objects that
collide?
6
Discuss how a un-magnetised ferromagnetic material
becomes magnetised.
7
Draw the magnetic field lines between two like poles
and two opposite poles
8
‘Denser (closer) field lines indicate a relatively stronger
magnetic field.’ Is this statement true, or is it false?
9
What rule can be used to determine the direction of
the magnetic field surrounding a current-carrying wire?
10 The diagram below shows a loop carrying a current
I that produces a magnetic field of magnitude B
(figure on the left). It is placed in a region where there
is already a steady field of magnitude B (the same
magnitude as that due to I) directed into the page
(figure on the right). The resultant magnetic field
inside the ring has a magnitude of 2B.
x
x
x
x
x
x
x
x
x
I
I
x x x x x
x x x x x
x x x x x
x x x x x
a What would the magnitude and direction of
the resultant field be at the centre of the loop if
the current in the loop is switched off?
b What would the magnitude and direction of
the resultant field at the centre of the loop be if
the current in the loop were doubled?
c What would the magnitude and direction of the
resultant field at the centre of the loop be if the
current in the loop were reversed but maintained
the same magnitude?
11 Describe how the magnetic fields of two currentcarrying wires parallel to each other interact, if the
current in both wires is running the same direction.
12 The figure below shows a cross-sectional view of a
long, straight, current-carrying conductor, with its axis
perpendicular to the plane of the page. The conductor
carries an electric current into the page.
A
N
D
B
C
W
E
S
a Using the right-hand grip rule, what is the direction
of the magnetic field produced by this conductor at
point B?
b If the current through the conductor were doubled,
what is the direction of the magnetic field produced
at point D?
c If the direction of the current were reversed and
halved, what is the direction of the magnetic field
produced at point A?
CHAPTER 14 | MAGNETISM
417
CHAPTER REVIEW CONTINUED
13 A current-carrying wire produces a magnetic field
in the form of concentric circles surrounding the
wire. How would the magnitude of the magnetic field
change when measured further away from the wire?
14 Which of the following statements is correct about the
magnetic field strength in a solenoid?
A The magnetic field at a point well inside a solenoid
is uniform and dependent on the length of the
solenoid.
B The magnetic field at a point well inside a solenoid
is non-uniform and dependent on the length of the
solenoid.
C The magnetic field at a point well inside a solenoid
is uniform and independent of the length of the
solenoid.
D The magnetic field at a point well inside a solenoid
is non-uniform and independent of the length of the
solenoid.
15 A solenoid carrying a current of 10 A produces a
magnetic field of 6 × 10−4 T.
a If the current was increased to 40 A and the number
of turns per unit length was reduced by a factor
of two, what would the resulting magnetic field
strength be?
b If the current was decreased to 5 A and the number
of turns per unit length was reduced by a factor
of two, what would the resulting magnetic field
strength be?
c If the magnetic field remained the same, and the
number of turns was increased by a factor of 10,
how much would the current required to generate
the magnetic field be reduced by?
418
MODULE 4 | ELECTRICITY AND MAGNETISM
16 Imagine you are making an electromagnet, and you
want to create a strong magnet that uses the least
amount of power. What would be the most appropriate
method to increase the magnetic strength while using
the least amount of power?
17 A current-carrying wire runs horizontally across a
table. The conventional current direction, I, is running
from right to left. The current-carrying wire produces
a magnetic field of 2.8 × 10−6 T measured 8.0 cm from
the wire.
a What current must the wire be carrying to produce
that field?
b Draw a diagram showing the direction of the
magnetic field around the wire.
18 A current-carrying wire produces a magnetic field of
5.9 × 10−6 T. Calculate the distance from the wire this
was calculated at if the current is known to be:
a 3A
b 10 A.
19 A solenoid produces an electric field of 7.2 × 10−6 T. It
has 100 turns and is 0.5 m long. Calculate the current
within the wire.
20 An electromagnetic lock, or magnetic lock, is a
device that consists of an electromagnet. Using your
knowledge of electromagnets, describe how this
locking mechanism works.
21 After completing the activity on page 404, reflect
on the inquiry question: How do magnetised and
magnetic objects interact? Qualitatively describe
how the metal rod was affected by the induced
magnetic field.
MODULE 4 • REVIEW
REVIEW QUESTIONS
MR
4
Electricity and magnetism
Multiple choice
1
2
3
4
When a voltmeter is used to measure the potential
difference across a circuit element, where should it
be placed?
A in series with the circuit element
B in parallel with the circuit element
C either in series or in parallel with the circuit element
D neither in series nor in parallel with the circuit
element
When an ammeter is used to measure the current
through a circuit element, where should it be placed?
A in series with the circuit element
B in parallel with the circuit element
C either in series or in parallel with the circuit element
D neither in series nor in parallel with the circuit element
Which one or more of the following statements is true
when two circuit elements are placed in series?
A The power produced in each element will be the same.
B The current in each element will be the same.
C The voltage across each element will be the same.
D The resistance of the combination will double.
Which one or more of the following statements is true
when two circuit elements are placed in parallel?
A The power produced in each element will be the same.
B The current in each element will be the same.
C The voltage across each element will be the same.
D The resistance of the combination will halve.
The following information applies to questions 5 to 8.
Three identical light-emitting diodes (LEDs) L1, L2 and L3,
are connected into the circuit as shown in the diagram
below.
3.0 V
WS
4.10
WS
4.11
5
When switch S is closed, the brightness of L1 will:
A be greater than previously
B be less than previously but greater than zero
C remain the same
D be equal to zero
6
When switch S is closed, the brightness of L3 will:
A be greater than previously
B be less than previously but greater than zero
C remain the same
D be equal to zero
7
When switch S is closed, the reading on voltmeter
V will:
A be greater than previously
B be less than previously but greater than zero
C remain the same
D be equal to zero
8
When switch S is closed, the power output of the
battery will:
A be greater than previously
B be less than previously but greater than zero
C remain the same
D be equal to zero
The following information applies to questions 9 to 11.
In the circuit shown, the current through R1 = I1 and the
current through ammeter A = IA. The voltmeter V has
extremely high resistance and ammeter A has negligible
resistance.
V
R1
X
R2
Y
V
L1
L3
A
L2
9
S
All three LEDs are operating normally. Assume that all
connecting wires have negligible resistance.
Which of the following statements is true?
A I1 = IA
B I1 > IA
C I1R1 = IA(R1 + R2)
D I1R1 = IAR2
REVIEW QUESTIONS
419
MODULE 4 • REVIEW
10 If another resistor R3 was connected between X and Y,
what would happen to the meter readings?
A The reading on V would increase; the reading on A
would decrease.
B The reading on V would decrease; the reading on A
would increase.
C They would both decrease.
D They would both increase.
16 Calculate the magnitude of the acceleration of an
electron in a uniform electric field of 6.7 N C−1.
The mass of an electron is 9.11 × 10−31 kg and its charge
is −1.602 × 10−19 C.
A 1.07 × 1019 m s−2
B 10.72 × 1019 m s−2
C 11.8 × 1012 m s−2
D 11.8 × 1011 m s−2
11 If another resistor R3 was connected between X and Y,
what would happen to the power output of the battery?
A It would increase.
B It would decrease.
C It would remain the same.
17 Which one or more of the following diagrams describes
the strongest, uniform magnetic field?
12 If a person appears to have been electrocuted, what
should you do first?
A Pull them away from the electrical device.
B Call 000.
C Commence CPR immediately.
D Switch off the power.
13 Which of the following options correctly describes the
electrostatic force?
A A contact force created by a charged particle.
B A non-contact force produced by an electrically
charged object.
C A non-contact force produced by a solenoid.
D A contact force produced by a metal rod with an
electric charge.
14 A student constructs a simple DC electric motor
consisting of N loops of wire wound around a wooden
armature, and a permanent horseshoe-shaped magnet
with a magnetic field of strength B. The student
connects the motor to a 9 V battery but is not happy
with the speed of rotation of the armature. Which one
or more of the following modifications will most likely
increase the speed of rotation of the armature?
A Increase the number of turns N.
B Use a 12 V battery instead of a 9 V battery.
C Replace the wooden armature with one of soft iron.
D Connect a 100 Ω resistor in series with the armature
windings.
15 The magnitude of an electric field surrounding a positive
point charge (q = 3.0 × 10−6 C) is found to be
1.32 × 107 N C−1. What distance from the point charge
was this calculated at? (Use k = 9 × 109 N m2 C−2.)
A 3.5 cm
B 4.0 cm
C 4.5 cm
D 5.0 cm
420
MODULE 4 | ELECTRICITY AND MAGNETISM
A
C
× × × ×
× × × ×
× × × ×
×
×× ×
××
×
×
××
× ×× ×
B
D
× × × ×
× × ×
× × ×
× ×
× × × ×
× × × ×
18 What is the approximate magnitude of the magnetic
field 1.1 cm from a 20 mA current-carrying wire?
A 2 × 10−6 T
B 3 × 10−6 T
C 4 × 10−6 T
D 5 × 10−6 T
19 How can you tell the strength of a magnetic field by
looking at a field diagram?
A from the density of the field lines
B from the direction of the field lines
C from the number of points where the field lines cross
D when field lines are evenly spaced
20 In the right-hand grip rule, in what direction do your
fingers point?
A the direction of the current
B the direction of the magnetic field
C the direction of the electric field
D the direction of the force
Short answer
21 Birds can sit unharmed on a high-voltage power line.
However, on occasion large birds such as eagles are
electrocuted when they make contact with more than
one line. Explain.
22 A student needs to construct a circuit in which there is a
voltage drop of 5.0 V across a resistance combination, and
a total current of 2.0 mA flowing through the combination.
She has a 4.0 kΩ resistor that she wants to use and
proposes to add another resistor in parallel with it.
a What value should she use for the second resistor?
b Determine the effective resistance of the combination.
23 A light bulb rated as 6.0 W and 0.50 A is to be operated
from a 20 V supply using a limiting resistor, R. The
circuit is represented below.
20 V
its north end towards the electromagnet is placed to the
right of it. The switch is initially open.
a Describe the force on the bar magnet while the switch
remains open.
b Describe the force on the bar magnet when the
switch is closed and a heavy conventional current
flows through the electromagnet from left to right.
c The battery is removed and then replaced so that the
current flows in the opposite direction. Describe the
force on the bar magnet now when the switch is closed.
30 A test charge is placed at point P, 30 cm directly above
a charge, q of +30 × 10−6 C. What is the magnitude and
direction of the electric field at point P?
R
6W
0.5 A
a What voltage is required to operate the light bulb
correctly?
b How many volts will there be across R when the
circuit is operating correctly?
c What is the required resistance of R?
Extended response
31 The following figure describes a simple electric circuit
in which a length of resistance wire is connected to a
battery (Use qe = 1.60 × 10−19 C).
R = 50 Ω
24 Explain whether the devices in a household circuit are
connected in series or in parallel and give a reason for
this wiring.
25 Two charges of +5 μC and −7 μC are positioned 0.4 m
apart in air. What is the force that acts between them?
27 The left diagram below represents two conductors, both
perpendicular to the page and both carrying equal
currents into the page (shown by the crosses in the
circles). In these questions ignore any contribution from
the Earth’s magnetic field. Choose the correct options
from the arrows A–D and letters E–G.
A
D
Q
P
×
R
B
×
C
E out of page
F into page
G zero field
What is the direction of the magnetic field due to the
two currents at each of the following points?
a point P
b point Q
c point R
28 A solenoid produces an electric field of 6.8 × 10−6 T. It
has 50 turns and is 0.6 m long. Calculate the current
within the wire.
29 An electromagnet with a soft iron core is connected in
series to a battery and a switch. A small bar magnet with
+ 100 V
a How many electrons pass through the wire every
second?
b How much electrical energy does each electron lose
as it moves through the wire?
c What happens to the electrical energy of the electrons
as they move through the wire?
d How much power is being dissipated in the resistance
wire?
e What is the total energy being supplied to all the
electrons passing through the wire each second?
f What is the power output of the battery?
g Discuss the significance of your answers to parts d
and f.
32 A special type of circuit device X has the I–V
characteristic depicted in the following graph.
2.5
2.0
I (mA)
26 Two parallel plates have a distance of 3.8 cm between
them and have a potential difference of 400 V across
them. What is the size of the electric field strength
between the plates?
1.5
1.0
0.5
20
40
60 80
V (V)
90 100
REVIEW QUESTIONS
421
MODULE 4 • REVIEW
When circuit device X is connected into the circuit, as
shown in the circuit diagram below, the reading on the
ammeter is 2.0 mA.
X
R = 50 kΩ
V1
V2
I
mA
250 V
a How would you describe the device X and what is its
purpose?
b Describe the resistance of such a device in the
voltage range 60−100 V.
c What are the readings on voltmeters V1 and V2?
d How much power is being consumed by the circuit
device X?
e How much power is being consumed by the resistor?
f What is the power output of the 250 V battery?
33 a Explain what is meant by the term potential
difference.
b Two parallel plates have a potential difference
of 1000 V and are 0.020 m apart. Calculate the
magnitude of the electric field.
c A gold(III) ion is accelerated by the electric field
created between the two parallel plates. The ion
carries a charge of +3qe and has a mass of
3.27 × 10−25 kg. Calculate the work done by the field
to move the ion.
(Use qe = −1.602 × 10−19 C.)
422
MODULE 4 | ELECTRICITY AND MAGNETISM
d If the ion starts from rest, calculate its final speed.
e What potential difference is needed across the plates
if you were to accelerate the ion up to a velocity of
6.5 × 104 m s−1?
f Using the answer to part e, calculate the force on the
charge as it travels through the electric field.
34 A series circuit is connected to a 9 V battery.
a A 5 kΩ light bulb is connected to the circuit. What is
the current through the lightbulb?
b A second identical light bulb is connected in series.
What is the voltage through each of the lightbulbs?
c A third light bulb is connected in parallel to the
second bulb attached in part b. What is the voltage
through each of the three bulbs?
d What is the ratio of power from the first to the third
light bulbs?
35 a A simple point charge with q = qe creates an electric
field. Calculate the strength of the field at a distance
of 10 cm.
b A second charge with q = +3qp is placed 2 cm from
the first charge. What is the force induced between
the two charges?
c Two parallel charged plates with a potential
difference of 800 V are separated by a distance of
20 cm. What is the electric field strength between the
plates?
d The charge (q = 3qp) travels between the two plates.
Calculate the force on the charge as it travels
through the electric field.
Answers
Chapter 1 Working scientifically
3
2.5
1.1 Questioning and predicting
2
3
4
5
1.2 Planning investigations
1
2
3
4
5
6
B
the type of grip (i.e. either two-handed or one-handed)
a In a controlled experiment, two groups of subjects are tested.
The groups, or the tests performed on them, are identical
except for a single factor (the variable).
b The dependent variable is the variable that is measured to
determine the effect of changes in the independent variable
(the variable that is changed). For example, in an experiment
testing the effect of soil pH on flower colour, the independent
variable would be soil pH and the dependent variable would be
flower colour.
a A stopwatch is often able to measure time to the nearest tenth
or hundredth of a second, which could give very precise results,
but accuracy depends on your reaction time, so errors could be
introduced.
b A clock can be used to measure time to the nearest second,
which is less accurate than the stop watch. Again, it relies on
your reaction time, which could introduce errors.
c Using a camera to record the motion means you can calculate
a more accurate value for time taken.
Using a ladder or standing at a height to drop the balls may be
a risk.
a valid b reliable c accurate
1.3 Conducting investigations
1
2
3
4
5
6
2
a A hypothesis is a statement that can be tested. This involves
making a prediction based on previous observations.
b A theory is a hypothesis supported by a great deal of evidence
from many sources. A principle is a theory that is so strongly
supported by evidence that it is unlikely it will be shown to
be untrue.
B
a if voltage is measured in units of number of batteries
b if voltage is measured with a voltmeter
qualitative (ordinal)
C
A: random error, B: systematic error
a systematic error b random error
15 m s−1
2 significant figures
• a lack of objectivity
• a lack of clear and simple instructions
• a lack of appropriate equipment
• a failure to control variables.
• a poor hypothesis that could not be tested objectively
• conclusions that do not agree with the results
• interpretations that are subjective.
a systematic error b random error
1.4 Processing data and information
Voltage (V)
1
1.5
1
0.5
0
4
5
6
0
0.02
0.04
Current (A)
0.06
0.08
Add a trend line or line of best fit.
31 ± 5.6
a ± 0.5ºC b ± 2.5%
1.5 Analysing data and information
TY 1.5.1 F = 9.9m − 0.06 TY 1.5.2 0.04 W
1 proportional relationship
2 inversely proportional relationship
3 directly proportional
4 time constraints and limited resources
5 9.6 represents the acceleration of the ball in ms −2, and
1.3 represents its initial speed in ms −1.
1.6 Problem solving
1
2
3
4
B
Higher mass coincides with lower acceleration, which supports the
hypothesis.
Different objects fell at different speeds. This refutes the
hypothesis. You may conclude that:
• The investigation needs more data in order to find a trend, or
your method of finding the results is unreliable.
• The investigation possibly did not factor in air resistance, and
should be repeated with objects with different masses but the
same air resistance.
‘Thirty repeats of the procedure were conducted’ is better because
the number of trials is quantified.
1.7 Communicating
1
3
4
5
B
2
D
a m s–2 b N m c kg m s–1 d kg m–3
4500 MW
When using different formulas or comparing results, all variables
need to be in the same units or there will be errors in your analysis.
Chapter 1 review
1
2
TY 1.4.1 49 ± 4.4 m s
1 An outlier is a data point that does not fit the trend.
2 a 23 b 19 c 22.5
−1
3
4
A hypothesis is testable prediction, based on evidence and prior
knowledge, to answer the inquiry question. A hypothesis often
takes the form of a proposed relationship between two or more
variables.
2 Form a hypothesis.
5 Collect results.
3 Plan experiment and equipment.
7 Draw conclusions.
6 Question whether results support hypothesis.
1 State the inquiry question to be investigated.
4 Perform experiment.
elimination, substitution, isolation, engineering controls,
administrative controls, personal protective equipment
dependent variable: flight displacement, independent variable:
release angle, controlled variable: (any of) release velocity, release
height, landing height, air resistance (including wind)
ANSWERS
423
6
8
9
10
11
12
14
15
16
17
18
19
20
21
22
23
a the acceleration of the object
b the vertical acceleration of the falling object
c the rate of rotation of the springboard diver
6.8 ± 0.4 cm s −1
7
the mean
non-linear relationship
a straight line with a positive gradient
Any issues that could have affected the validity, accuracy, precision
or reliability of the data plus any sources of error or uncertainty.
Bias is a form of systematic error resulting from a researcher’s
personal preferences or motivations.
C
13
2500 μm
Any of the following are correct:
• the activity that you will be carrying out
• where you will be working, e.g. in a laboratory, school grounds,
or a natural environment
• how you will use equipment or chemicals
• what clothing you should wear.
a i validity
ii reliability
b The conclusion is not valid because it does not relate to the
original hypothesis.
c To improve the validity of the experiment, the hypothesis
should specify the group that is being studied, i.e. teenage
boys. A larger sample size should be used to improve reliability.
The group must be divided into an experimental group and
control group. The experimental group should only eat fast food
and the control group should eat a normal diet. To improve
accuracy all measurements should be performed using the
same calibrated equipment. Using different equipment can also
assist in eliminating systematic errors. Repeat readings of the
liver function test should be performed on each sample and the
average calculated to minimise the effects of random errors.
To avoid plagiarism and ensure creators and sources are properly
credited for their work.
± 0.05%
a a variable which must be kept constant during an investigation
b a separate practical investigation that is conducted at the same
time where the independent variable is kept constant.
qualitative: robust aroma, frothy appearance, strong taste, white
cup; quantitative: cost $3.95, coffee temperature 82°C, cup height
9 cm, volume 180 mL
A trend is a pattern or relationship that can be seen between the
dependent and independent variables. It may be linear, in which
the variables change in direct proportion to each other to produce
a straight trend line. The relationship may be in proportion but
non-linear, giving a curved trend line. The relationship may also be
inverse, in which one variable decreases in response to the other
variable increasing. This could be linear or non-linear.
Accuracy refers to the ability of the method to obtain the correct
measurement close to a true or accepted value. Validity refers
to whether an experiment or investigation is actually testing the
hypothesis and aims.
a bar graph b line graph
c scatter graph (with line of best fit) d pie chart
a ± 0.5°C b ± 2.4% c 3
Chapter 2 Motion in a straight line
2.1 Scalars and vectors
TY 2.1.1 a 50 N west
b −50 N
TY 2.1.2 19 N down
TY 2.1.3 1988 m s −1 up
1 4 m west
2
2 m down
3
11 m forwards
4 D
5
8 m s −1 east
6
2 m s −1 left
7 7 m s −1 downwards
8 14.0 m s −1 backwards
2.2 Displacement, speed and velocity
TY 2.2.1 a 0.92 m s −1 east
1 B and C
424
ANSWERS
b 3.3 km h −1 c 4.0 m s −1
d 14 km h −1
2
3
4
5
6
7
8
a
c
a
a
c
a
a
a
a
c
s = +40 cm d = 40 cm b s = −10 cm d = 10 cm
s = +20 cm d = 20 cm d s = +20 cm d = 80 cm
80 km b +20 km or 20 km north
−10 m or 10 m downwards b +60 m or 60 m upwards
70 m d +50 m or 50 m upwards
33 m s −1 b 25 m
17 km h −1 b 4.7 m s −1
0.9 m s −1 b 0.1 m s −1 east
21 km b 15 km north
14 km h −1 d 10 km h −1 north
2.3 Acceleration
TY 2.3.1 a −2.0 m s −1 b 16 m s −1 upwards
TY 2.3.2 460 m s−2 upwards
1 −7 km h −1
2
5 m s −1 upwards
3 8 m s −1 upwards
4
5.0 m s−2 south
5
43 km h −1 s −1
6 a −10 m s −1 b 40 m s −1 west c 800 m s−2
7 a 8.0 m s −1 b 8.0 m s −1 south c 6.7 m s−2, south
2.4 Graphing position, velocity and acceleration
over time
TY 2.4.1 a 15 m s −1 backwards
b the cyclist is not moving
TY 2.4.2 a 4 m west b 2 m s −1 west
TY 2.4.3 2 m s−2 west
1 D
2 The car initially moves in a positive direction and travels 8 m in 2 s.
It then stops for 2 s. The car then reverses direction for 5 s, passing
back through its starting point after 8 s. It travels a further 2 m in a
negative direction before stopping after 9 s.
3 Reading from graph:
a +8 m b +8 m c +4 m d −2 m
4 t = 8s
5 a +4 m s −1 b 0 c −2 m s −1 d −2 m s −1 e −2 m s −1
6 a 18 m b −2 m
7 a 2 m s −2 b 10 s c 80 m d 7 m s −1 forwards
8 a
Acceleration (m s–2)
5
2
1
0
0
2
4
6
8 10 12 14 16 18 20
Time (s)
b +12 m s −1
2.5 Equations of motion
TY 2.5.1 a 3.8 m s−2 west b 4.0 s c 7.5 m s −1 east
1 E
2 a 3.1 m s−2 forward b 50 m s −1 c 180 km h −1
3 a +2.0 m s−2 b +8 m s −1 c 64 m
4 a 40 m s−2 upwards b 1120 m c 580 km h −1
d 80 m s −1 e 124 m s −1
5 a 5.0 m s−2, east b 2.7 m c 5.5 m s −1
6 D
7 a −2.4 m s−2 downwards b 1.8 s c 3.1 m s −1 downwards
8 a 21 m s −1 b 5.2 m c 36 m d 41 m
2.6 Vertical motion
TY 2.6.1 a 2.5 s b 3.5 s c 34 m s −1 downwards
TY 2.6.2 a 11.48 m b 3.06 s
1 B
2 A and D
3 a 29 m s −1 b 24 m s −1 c 12 m s −1 downwards
4 a 15 m s −1 upwards b +11.0 m
5 a 3.9 m s −1 b +0.78 m c 0.20 m d 0.58 m
6 a 2.0 s b 20 m s −1 c 20 m d –20 m s −1
Chapter 2 Review
3.3 Relative motion
1
3
TY 3.3.1 89 km h–1 N52°W
TY 3.3.2 4.3 m s–1 S28°W
TY 3.3.3 904 km h–1 S84.6°E
1 vCT = vCG − v TG
2
2.6 m s–1 N17.7°W
3 5.4 m s–1 N68°E
4
N9.46°E
5 a 840 km h–1 south b 913 km h–1 S4.40°E
6 4 km h–1 towards the finish line
7
21 m s–1 N73°W
4
5
6
7
B and D
2
A and D
The vector must be drawn as an arrow with its tail at the point of
contact between the hand and the ball. The arrow points in the
direction of the ‘push’ of the hand.
Vector A has twice the magnitude of B.
Signs enable mathematical calculations to be carried out on the
vectors.
+80 N or just 80 N
The resultant force vector is 5 N.
Chapter 3 review
1
2
3
34.0 m s−1 north and 12.5 m s−1 east
70° anticlockwise from the left
a
N
25º
−5 −4 −3 −2 −1
8
11
13
14
17
18
19
20
21
22
23
24
25
26
27
28
29
0
1
2
3
4
5
6
7
force (N)
21.0 m backwards
9
6 m s −1 left
10
26 m s −1
54 km h −1
12
15 km h −1
a 10 km h −1 north b 2.8 m s −1 north
−2.0 m s −1
15
B
16
–6 m s−2
a from 10 to 25 s b from 30 to 45 s
c from 0 to 10 s, from 25 to 30 s and from 45 to 60 s
d 42.5 s
a B b A c C
a 114 m north b 10.4 m s −1 c 0 m s−2
d 7 m s−2 south e A
16 m s −1
a 4.0 m s−2 b 4.0 m s −1
c 6.0 m
a −5.0 m s−2, forwards b 2.0 s
a +4 m b A and C c B, +0.8 m s −1 d D, 2.4 m s −1
e 0.80 m s −1
a 8.0 s b 16 s c 192 m
a 3.5 s b 2.9 s
a 1.7 s b 3.2 s
a 4.0 m s −1 b 5.7 m s −1 c 2.0 s d 0.85 s.
The second bolt is 0.31 m away from the first.
The third bolt is 0.92 m away from the second.
The fourth bolt is 1.5 m away from the third.
Responses will vary. The equations of motion are a helpful tool to
describe the straight-line motion of an object moving with constant
acceleration. In the activity, the object is falling under constant
acceleration (gravity), so the time each bolt takes to fall can be
predicted. Graphs of velocity versus time and acceleration versus
time can be helpful when describing an object’s motion.
Chapter 3 Motion on a plane
3.1 Vectors in two dimensions
W
TY 3.2.1 3168 N left, 1580 N downwards
1 a 265 N downwards b 378 N right
2 19.8 N south and 16.6 N east
3 4.55 m s−1 north and 17.7 m s−1 west
4 18.9 m south and 43 m east of his starting point
5 109 000 N north and 208 450 N west
6 a 50 N south, 87 N east b 60 N north
c 282 N south, 103 N east
d 1.0 × 105 N upwards, 2.6 × 105 N horizontally
S
N
b
74º
W
4
5
8
9
10
12
14
16
17
19
E
S
a sin b sin c tan d cos
8.5 m
6
2.4 m s–1 west, 6.0 m s–1 north
7
C
horizontal component 150 N, vertical component 260 N
C
4.5 m s–1 horizontally, 2.6 m s–1 vertically
11
6.7 m
vPH = vPG − vHG
0.85 m s–1 N69°E
13
42 km h–1
15
8.3 m s–1 N15°W
a north b 930 km h–1 north
3.8 m s–1 N39°W
18
120 km h–1 N59°E
–1
–1
a 0.14 m s east, 0.79 m s north
b The vector components show the relative velocity of the wind
to the table and the marble to the wind, i.e. they describe
vMT = vMW + v WT
20 a
TY 3.1.1 50° clockwise from the right direction
TY 3.1.2 5.8 N N59°E
TY 3.1.3 9.2 m s−1 N41°E
1 533 m s−1 N49.6°W
2
59.4 m s−1 N45.0°W
3 8.79 m s−1 N36.7°W
4
45 m, S63°W
5 6325 N N71.57°E
3.2 Vector components
E
1.5 m s–1
ν→
4.0 m s–1
edge of the river
b
c
d
e
21 a
4.3 m s–1 at 69° from edge of river
1300 m at 69° from edge of river
1200 m in the forward direction of the boat
450 m downstream
A ball thrown through the air experiences the forward force
from the initial throw, the force due to air resistance, and the
downwards weight force due to gravity. The horizontal and
vertical motion is slowed by air resistance, and the weight force
pulls the ball down.
b The ball is a projectile; and all projectile motion can be
described by a parabola.
ANSWERS
425
22 Responses will vary. Motion in two and three dimensions can
be analysed by considering the perpendicular components
independently. In this experiment the original motion (along
the table) and the direction of the fan (across the table) are
perpendicular. The fan provides a force on the marble across the
table, causing an acceleration, so the component of the velocity in
this direction increases. This is similar to gravitational acceleration
in the vertical component of projectile motion. The component
along the table should remain relatively unchanged. There will be
some friction, slowing the motion in this direction. This is similar
to the horizontal component of projectile motion. Each component
can be analysed using the equations of motion to predict how an
object’s velocity and displacement will change over time.
c
1
0.9
0.8
Acceleration (m s–2)
0.7
0.6
0.5
0.4
Module 1 review
1
3
4
9
10
11
16
21
D
a
B
a
a
C
C
a
c
22 a
23 a
c
24 a
25
26
27
28
b
d
a
a
a
a
2
A
b B
5
A
6
C
7
C
8
D
C b B
A, B, C b A, E
12
D
13
B
14
A
15
D
17
D
18
A
19
A
20
C
10.2 s b 510 m
9.8 m s−2 downwards
13.9 km N21°E b 0.6 m s−1
5 s b 4.07 s
platinum sphere: .24.4 m s−1, lead sphere: 30.0 m s−1 downwards
For the first 30 s the cyclist travels east 150 m at a constant
speed. Then they accelerate for the next 10 s, travelling 150 m.
They then travel at a higher constant speed for the next 10 s,
travelling another 200 m.
5 m s−1 east c 20 m s−1 east
15 ms east e 1.5 m s−2 east f 10 m s−1
deceleration of 0.69 m s−2 b 2.5 × 102 m
2.7 s b −5.6 m s−2
10 m b 5.0 s c 60 m
From the starting point (the origin), the person walks forward
in a positive direction for 2.5 m in 2 s, and then stays in that
position for 4 s more. Then they turn and walk back in a
negative direction, passing the origin, to 2.5 m behind the
starting point over 2 s. They stay at this point for 2 s before
walking back to the origin.
b
Velocity (m s–1)
0.2
0.1
0
29
30
31
32
33
34
0
2
4
6
a
2
35 a 105 km h−1 N64°E
1
0.5
–2
–2.5
ANSWERS
2
a
b 105 km h−1 S64°W
4.1 Newton’s first law
–1.5
14
2as = 2(uv − u ) + v 2 − 2uv + u2 ; v 2 = u + 2as
Chapter 4 Forces
–1
12
2
2
2
1.5
–0.5
10
2
c s = ut + 1 at 2; s = u  v − u  + 1  v − u  ;
2
0
8
Time (s)
Although accelerations (or decelerations) must occur at 0, 2, 6, 8,
10 and 12 seconds because the velocity changes at these times,
they are more or less instantaneous and therefore cannot be
shown on the graph.
a 10.4 m s−1 b 1.6 minutes
a 1.4 s b 14 m s−1
a i 0.10 m s−1 ii 0.30 m s−1 iii 0.50 m s−1
b i 0.10 m s−1 ii 0.30 m s−1 iii 0.50 m s−1
c constant acceleration
a 34 km h−1 b 567.6 m
c i 22.65 km h−1 S76.2°E ii 31.4 km h−1 south
d 1.1 km
a 4.4 m s−1 N35°E b 3.65 m s−1 N9.5°W
c 219 m N9.5°W d yes, up to rounding error
a a = ∆v ; v − u = at ; v = u + at
t
b vav = 1 (v + u); vav = 1 (2u + at ); vt = ut + 1 at 2; s = ut + 1 at 2
2
3
2.5
426
0.3
C
2
4
6
8
10
12
14
Time (s)
c 157 km
TY 4.1.1 net force 3 N right; 3 N to left required for equilibrium
TY 4.1.2 a 13 N at 22.6° to left of vertically down
b 13 N at 22.6° to right of vertically up
1 The box has changed velocity, so can use Newton’s first law to
conclude that an unbalanced force must have acted to slow
it down.
2 The car has maintained its speed but its direction has changed, so
velocity has changed. From Newton’s first law, can conclude that
an unbalanced force has acted on the car.
3 B
4 When the bus stops suddenly the passenger will continue to move
with constant velocity unless acted on by an unbalanced force
(Newton’s first law of motion); no force pushes them forward.
5 The plane slows down very rapidly as it travels along the runway.
A passenger standing in the aisle would have no retarding forces
acting and so would maintain their original velocity and move
rapidly towards the front of the aeroplane.
6
7
8
a The inertia of the glass makes it remain at rest. The cloth is
pulled quickly so the force of friction between the cloth and
glass acts only for a very short time, not long enough to enable
friction to overcome the inertia of the glass and make it move.
b Using a full glass makes the trick easier because the inertia of a
full glass is greater than that of an empty glass.
lift force 50 kN upwards, drag 12 kN west
net force 5 N to right; for equilibrium need 5 N to left
b
→
F2
→
F1
4.2 Newton’s second law
TY 4.2.1 307 N south
TY 4.2.2 2.50 m s−2 forwards
TY 4.2.3 150.6 N upwards, 2.0 m s−2 upwards
1 6.61 m s−2 north
2
9.80 m s−2 down
3 9.80 m s−2 down
4
78 500 kg
5 a 147 N down b 1.5 N north c 0.1 m s−2 north
d 7.2 N north
6 4 boxes
4.3 Newton’s third law
TY 4.3.1 a action force: Fby ball on floor; reaction force: Fby floor on ball
b
→
Fequilibrium
8
11
12
13
14
15
18
19
21
3
4
5
6
7
force
on hammer by nail, force on nail by hammer
a F by the Earth on the astronaut
b F by the astronaut on the Earth
force on her hand by water
a 140 N in opposite direction to fisherman
b 3.5 m s−2 in opposite direction to fisherman
c speed of fisherman 1.0 m s−1, speed of boat: 1.8 m s−1
away from ship
Tania. For an action−reaction pair, the two forces act on different
objects. In this case, both weight force and normal force act on the
lunch box.
49 N perpendicular to the plane
Chapter 4 review
1
2
3
4
5
6
7
a A and C
No. The passengers have inertia, so their masses resist the
change in motion as the train starts moving forwards. According
to Newton’s first law, they maintain their original state of being
motionless until an unbalanced force acts to accelerate them.
C
a 5 N to the right b no; net force is not zero
An internal force has no effect on the motion of the object,
whereas an external force does.
35 N in the positive x direction, 35 N in the positive y direction
a
→
F3
→
F2
B
9
38.3 kg
10
1560 m s−2 south
5.40 m s−2 north
119 N, 11.9 m s−2, both in the positive x direction
19 N, 0.25 m s−2, both in the positive y direction
The bus will accelerate more slowly than the motorcycle because
an object’s acceleration is inversely proportional to the object’s
mass (Newton’s second law).
29 N
16
10.2 m s−2 upwards
17
C
The compressed air inside the balloon escapes from the mouth of
the balloon. The escaping air exerts an equal and opposite forwards
force on the balloon, which moves the balloon around the room.
100 N east
20
75.0 N north
Responses will vary. Forces are either contact forces or forces
mediated by a field. Contact forces occur when objects come into
contact, e.g. a racquet hitting a ball. Non-contact forces act on an
object in a field, e.g. the gravitational force keeping the Moon in
orbit around Earth. An object remains at rest (or constant velocity)
unless an unbalanced force acts on it (Newton’s first law). It is in
equilibrium, so the sum of all forces acting on it are equal to zero.
Using the concept of equilibrium and Newton’s second law you
can predict an object’s motion by analysing the forces acting on it.
Chapter 5 Forces, acceleration
and energy
5.1 Forces and friction
TY 5.1.1 50 N in opposite direction to motion of wardrobe
TY 5.1.2 The puck will slow down and stop sliding.
TY 5.1.3 a 0.49 N in opposite direction to motion of puck
b 4.9 m s−2 in opposite direction to motion of puck
TY 5.1.4 a 7.0 m s−2 forwards b 5.0 m s−2 forwards
TY 5.1.5 a
Newton’s second law experiment
6.0
5.0
Acceleration (ms–2)
1
2
→
F3
4.0
3.0
2.0
1.0
0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18
1/total mass (kg–1)
→
F1
→
Fnet
bnet force 33 N, expected net force 49 N; there is
systematic error
1 a decrease b decrease c increase d increase
2 a kinetic b kinetic c static
3 Without friction the car would follow Newton’s first law and travel
in a straight line at a constant speed. This would cause it to slide
across and off the road.
ANSWERS
427
4
5
6
incorrect calibration of instruments, zeroing error, friction,
air resistance, repeated errors in experimental method
(e.g. repeatedly failing to adjust for parallax error)
−20 N
a 1.6 m s−2 in the positive direction
b 0.2 m s−2 in the positive direction
5.2 Work
TY 5.2.1 250 J
1 Energy is the capacity to cause change whereas work occurs
when there is an energy change, e.g. when energy changes
from one form to another or is transferred from one object
to another.
2 a kinetic b potential c kinetic d kinetic e potential
3 10 000 J
4 The person exerts a force on the wall but the wall has no
displacement so no work is done.
5 18 N in the direction of motion
6 Since the box does not move, no work is done.
7 1.5 m
5.3 Energy changes
TY 5.3.1 78 J
TY 5.3.2 a −770 kJ b 19 kN
TY 5.3.3 114 km h−1
TY 5.3.4 15 J
TY 5.3.5 32 J
1 57 kJ
2
370 kJ
3
40 km h−1
4 doubling mass causes K to double also
5 a 4.6 J b 2.3 J
6
2360 kJ
5.4 Mechanical energy and power
TY 5.4.1 50 J
TY 5.4.2 3.8 m s−1
TY 5.4.3 76 m s−1
TY 5.4.4 700 W
TY 5.4.5 217 kW
TY 5.4.6 2600 W
TY 5.4.7 11 kW
1 a 26 000 J b 17 640 J
2
1.5 m
3 a 336 J b 29.0 m s−1
4 113 kW
5
24.5 kW
6 1800 N against the direction of motion
Chapter 5 review
1
2
3
6
10
13
16
19
20
22
23
24
26
27
25 N in direction of motion
Friction between the car’s tyres and the road is the unbalanced
force which is causing the car to decelerate. If there is almost no
friction, the car will not slow down; it will continue to travel in a
straight line at a constant speed.
1.5 m s−2
4
1.05 m s−2
5
90 N
+130 N
7
160 kJ
8
58.8 kJ
9
1.16 kJ
8.2 m
11
135 J
12
11 m s−1
by factor of 4
14
340 J
15
5.1 m
3.9 m s−1
17
4.1 m
18
14.4 m s−1
a 81 J b 81 J c 180 m s−1
196 kW
21
15.4 kW
1500 N against direction of motion
a 1920 J b 96 N in direction of motion
19.2 J
25 a 17 m s−1 b 14 m s−1
73.5 W
Responses will vary. When a force is applied to the hovercraft
over a distance, work is being done on the hovercraft, increasing
its energy. The larger the distance,
the larger the energy transfer,
according to the equation W = F s.
The kinetic energy of your push is transferred to the hovercraft.
Some energy will be lost to air resistance, but frictional forces
between the surface and the hovercraft are almost zero because
the balloon is not in contact with the surface. The distance
the hovercraft travels can be predicted by analysing the net
force acting on the balloon hovercraft, or by using the law of
conservation of mechanical energy.
428
ANSWERS
Chapter 6 Momentum, energy and
simple systems
6.1 Conservation of momentum
TY 6.1.1 20 500 kg m s−1 north
TY 6.1.2 0.80 m s−1 north
TY 6.1.3 1.56 m s−1 south
TY 6.1.4 1630 m s−1 south
TY 6.1.5 inelastic
1 8.75 kg m s−1 south
2
9610 kg m s−1 west
3 3.54 kg m s−1 south
4 second ball (17 kg m s−1 compared with 16 kg m s−1)
5 0.438 m s−1 backwards
6
0.93 m s−1 north
7 142 000 kg
8 3.0 m s−1 in direction opposite to exhaust gases
6.2 Change in momentum
TY 6.2.1 589 kg m s−1 south
TY 6.2.2 0.0512 kg m s−1 N38.7°E
1 83.1 kg m s−1 south
2
235 000 kg m s−1 east
3 40.0 kg m s−1 east
4
2.45 kg m s−1 down
5 2.4 m s−1 north
6
2860 kg m s−1 N45°E
7 377 kg m s−1 S42°W
6.3 Momentum and net force
TY 6.3.1 a 0.192 kg m s−1 upwards b 54.1 N upwards
TY 6.3.2 a 0.192 kg m s−1 upwards b 0.591 N upwards
TY 6.3.3 a +32 N b 0.36 kg m s−1 upwards
1 a 452 kg m s−1 east b 452 kg m s−1 east c 129 N east
2 Airbags increase the duration of the collision, which reduces
the rate of change of momentum of a person’s head during an
accident, decreasing the force, and so the severity of injury.
3 a 1.90 kg m s−1 east b 1.90 kg m s−1 east c 19.0 N east
d 6.34 N east
4 a +3.3 kg m−1 b 65 N in direction of ball
c 65 N in opposite direction to ball
5 Results for question 5 are estimates and may vary.
a 1200 N b 63 N s
6 a 1.25 kg m s−1 opposite in direction to initial velocity
b 1.25 kg m s−1 opposite in direction to initial velocity
c 1.6 × 103 N in opposite direction to initial velocity of arrow
7 a Helmet is designed so stopping time is increased by the
collapsing shell during impact. This reduces force, since
impulse = F = Δt = Δp
b no: would reduce stopping time and increase force
Chapter 6 review
6.7 × 104 kg m s−1
2
3.8 × 10−5 kg m s−1
chestnut
4
504 kg m s−1 west
221 kg m s−1 backwards
6
5.2 ms−1
0.558 m s−1 backwards
a +40 m s−1 b +4.5 × 103 N c 10.2 m s−2
1.6 m s−1 north
10
4.8 kg m s−1 away from racquet
480 kg m s−1 N38.7°E
12
520 kg m s−1 S55°W
0.58 kg m s−1 76° from the cushion
14
1.51 N east
10 kg m s−1
16
10 kg m s−1
17
59 m s−1
a 5.3 m s−1 b inelastic
a 30 kg m s−1 upwards b 300 N upwards
To reduce force on egg to less than 2 N, duration of collision
must be greater than 0.125 s. Cushion egg to increase duration
of collision, e.g. using bubble-wrap, cotton wool, pillow, pile of
sawdust or flour.
21 Responses will vary. The system in the activity is combined mass
of student and medicine ball. Momentum of an object depends
on mass and velocity. Objects with a larger momentum require
larger momentum transfer to come to a stop. Momentum transfer
requires a force to be applied for a period of time, the larger the
momentum transfer the larger the force or the longer the time. In
this inquiry you can feel the increased difficulty in stopping, and by
measuring the stopping distance you can quantifying that feeling.
1
3
5
7
8
9
11
13
15
18
19
20
Module 2 review
1
3
5
9
14
19
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
C
2
a D b A
D
4
A
C
6
A, B and C
7
A, B and D
8
A, B and C
D
10
C
11
C
12
D
13
A
B
15
C
16
D
17
C
18
B
A 20
C
a F EM is the gravitational force by Earth on the man; F N = F SM, the
normal
reactionforce exerted by the surface on the man
b F EM = 980 N = −F N (the forces are equal in magnitude but
opposite
in direction)
c F ME is the gravitational attraction the manexerts on the Earth,
which is the reaction force to his weight; F MS is the force that
the
man exerts on the surface.
a F EA is gravitational force exerted by Earth on A, directed
downwards (weight
force of the block); balanced by equal normal
reaction
force F BA directed upwards. Both forces are 100
N.
b F EB is weight of block B, which is 100 N downwards. F AB is force
exerted
by A on B, also 100 N downwards. The normal reaction
F
force
balances both, so it is 200 N upwards.
CB
c F BC is 200 N downwards
(caused by weight of A and B, and is
reaction pair to F CB), F EC = 100 N downwards and the normal
reaction
force F TC = 300 N is exerted
upwards by table on C.
d F EB is still exerted on block B, but F CB = 0. Both A and B fall, so
contact forces between A and B are also zero. Each block only
experiences its own weight force and accelerates under gravity.
a 600 N b 800 N c 4.0 m s−2 up the incline
a 3.3 m s−2 up for 10 kg mass and 3.3 m s−2 down for 20 kg mass
b 1.3 × 102 N
a 300 N to right b 1.0 kJ c 3.0 kJ d 3.0 kJ e 17.3 m s−1
a 140 J b 80 J c 6.4 m s−1 d 140 J.
a 1.1 × 104 N m−1
b just before striking trampoline on first descent
c gravitational potential energy ⇒ kinetic energy ⇒ elastic
potential energy of trampoline ⇒ kinetic energy as child
rebounds losing contact with trampoline (some loss to heat and
sound) ⇒ gravitational potential energy
Swimmer pulls against water with arms, exerting a force on the
water. The reaction force (Newton’s third law) of the water on her
arms propels her forward. If reaction force is greater than the sum
of drag forces on her, she will accelerate according to Newton’s
second law. If no net force, she will travel at constant velocity
(Newton’s first law).
Earth: 1.96 × 105 N, Moon: 3.2 × 104 N
a 645 kg m s−1 in positive direction
b Momentum is always conserved in a collision. The momentum
after the collision will be a combination of momentum of pole (as
it wobbles) and momentum of player as he rebounds off post.
c As the player’s head comes to rest against the pole, his brain
continues moving at 7.5 m s−1 (Newton’s first law) until it collides
with the inside of the skull, which could result in concussion.
a 1.6 MJ b 1.6 × 104 N forwards c 16.4 kN forwards
d 1.64 MJ e 320 kW f 40 kJ
a 40 J b 250 J
c 5.0 cm
d Elastic potential energy stored in spring is transferred back to
trolley as kinetic energy when spring starts to regain original
shape.
e It is elastic.
a from 2.0 to 5.0 cm; cube loses kinetic energy in this section
b 7.1 m s−1 c 3.0 J d converted into heat and sound
e 100 N
a +1.6 × 104 m s−2 b +8.8 × 106 N c +4.4 × 105 kg m s−1
d −4.1 m s−1 e −8.8 × 106 N
f W = 1.8 × 108 J; K = 1.8 × 108 J; represents an ideal situation;
realistically there would be significant losses
a 2.2 m s−1 b before 1.8 × 105 J; after 1.3 × 105 J
c inelastic; kinetic energy not conserved d 262 m
e −8.3 × 104 N
Chapter 7 Wave properties
7.1 Mechanical waves
1
2
3
4
5
Particles oscillate backwards and forwards or upwards and
downwards around a central position, but do move along with
the wave.
a false b true c true d true
Point B is moving downwards.
A has moved right and B has moved left.
C and D. Only energy is transferred by a wave.
7.2 Measuring mechanical waves
TY 7.2.1 amplitude = 0.02 m, wavelength = 0.4 m,
wavenumber = 15.7 m–1
TY 7.2.2 amplitude = 0.1 m, period = 0.5 s, frequency = 2 Hz
TY 7.2.3 7.5 × 1014 Hz
TY 7.2.4 1.3 × 10–15 s
1 a C and F b wavelength c B and D d amplitude
2 wavelength = 1.6 m, wavenumber = 3.9 m −1, amplitude = 0.2 m
3 a 0.4 s b 2.5 Hz
4 6.5 m s −1
5 a true b false c true d false
6 a λ = 4 cm, A = 0.5 cm b 0.02 m s −1 c red particle
7 5 × 10−6 s
Chapter 7 review
1
2
3
4
5
6
7
8
9
11
15
18
20
21
The particles on the surface move up and down as the waves
radiate outwards.
Similarities: both types of wave carry energy away from source
and are caused by vibrations.
Differences: in transverse waves particles vibrate at right angles to
direction of wave energy movement; in longitudinal waves particles
vibrate parallel to direction of wave energy movement.
In sound waves particles only move back and forth around an
equilibrium position, parallel to direction that wave energy travels.
They collide with adjacent particles and transfer energy, so they
cannot move along with the wave as they lose their kinetic energy
to the particles in front of them during the collisions.
The motion of the particles is at right angles to the wave’s
direction of travel.
longitudinal: a and d, transverse: b and c
Mechanical waves move energy via the interaction of particles.
The molecules in a solid are closer together than those in a gas. A
smaller movement is needed to transfer energy in a solid so the
energy of the wave is usually transferred more quickly.
The energy travels towards the left, away from the stone towards X.
U is moving down and V is momentarily stationary (and will then
move downwards).
period = 4 × 10−2 s, frequency = 25 Hz
10 5 ms
5 cm
12
2
13
2
14
B and D
0.300 m s −1
16
0.055 m
17
5m
1.1 m
19
4
Wavelength increases, velocity is constant
Responses will vary. Waves, both mechanical and electromagnetic,
can be quantified using frequency, wavelength and speed, using
wave equation v = f λ. Mechanical waves require a medium to
travel in while electromagnetic waves (such as visible light) do not.
The mechanical sound waves created by the tuning fork create
the water waves. It was possible to find the inverse relationship
between frequency and wavelength in this activity.
Chapter 8 Wave behaviour
8.1 Wave interactions
TY 8.1.1 0
1 wave is reflected, 180° change in phase
2 amplitude
3
B
4
52°
5
C
6 a faster in medium A b unchanged
7 diffraction through gap becomes less significant
ANSWERS
429
8.2 Resonance
2
3
4
5
An object subjected to forces varying with its natural oscillating
frequency will oscillate with increasing amplitude. This could
continue until the structure can no longer withstand the internal
forces and it fails.
B
D; same length as pendulum on left, so natural frequency of
pendulum D is equal to driving frequency of pendulum on left, and
maximum energy is transferred
Frequency of idling motor (100 Hz) is the forcing frequency in
this situation and must be equal (or close) to natural frequency
of vibration of mirror, so that maximum energy is transferred to
mirror and it vibrates with large amplitude. When truck is driving,
frequency of motor is no longer equal to natural frequency of
vibration of mirror, so amplitude of vibrations is much smaller.
Normal walking results in a frequency of 1 Hz or 1 cycle (i.e. two
steps) per second, and may result in increase in amplitude of
oscillation of bridge over time, which could damage it.
Chapter 9 Sound waves
9.1 Sound as a wave
1
3
Chapter 8 review
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
undergoes phase change
a transmission b reflection c absorption
a true
b false: As the pulses pass through each other, they interact
by either destructive or constructive superposition. Once the
pulses have passed each other they return to their original
shape.
c true
A
A incident ray, B normal, C reflected ray, D boundary between
media, E refracted ray
Bass sounds have larger wavelengths and diffract more readily, so
can spread out around obstacles and corners, making them easier
to hear.
As light passes from water into air its speed increases, and it
refracts away from the normal.
When viewed from directly above, the light ray strikes the interface
between the air and water at an angle of incidence of 0°, and no
refraction occurs. The fish will be observed at its actual position,
but will seem to be deeper in the water than in previous question.
The wavelength of light is too small to diffract through and around
everyday objects.
reflection and refraction
speed of sound in hotter air is faster
C
There would be no reflection or refraction, people in building may
find rock concert quieter
green wave
Must have same frequency (or wavelength) and same amplitude,
and must be out of phase by 180°.
All objects have a natural or resonant frequency. If it is made to
vibrate at this frequency by applying an outside, driving frequency,
the amplitude of the vibrations will increase as the maximum
possible energy is transferred. An example of resonance is in
musical instruments.
D
It is only the pattern made by the amplitude along the rope that
stays still at certain points. The rope still has waves traveling up
and down it. The waves superimpose but give the appearance of a
wave standing still at places, with parts of the rope oscillating up
and down.
The natural frequency of the human ear canal is approximately
2500 Hz. Since the driving frequency of the signal generator
(2500 Hz) is the same as this, there is maximum transfer of energy
to the ear, and this is perceived as a very loud sound.
All options are correct.
Responses will vary. As waves interact with an object, they cause
it to vibrate at certain frequencies. Objects tend to vibrate at a
certain frequency known as their resonant (or natural) frequency.
430
ANSWERS
B
a E
c
Pressure variation
1
If an outside source, such as the pull on the tin, causes the object
to vibrate at its natural frequency, resonance will occur. This
causes a maximum transfer of energy. This is shown in the activity
as a maximum amplitude of displacement.
4
2
A and D
b 0.40 m
0.20
0.40
0.60
Distance from source (m)
a 0.5, 2.5, 4.5 ms
0.80
b 500 Hz
9.2 Sound behaviour
TY 9.2.1 1.0 × 10−4 W m−2
TY 9.2.2 694 Hz
TY 9.2.3 345 Hz or 353 Hz
1 amplitude
2
1.0 W m−2
3 the same: speed of each vehicle is the same and there is no
relative motion of the medium
4 102 Hz
5
3 Hz
9.3 Standing waves
TY 9.3.1 a 0.50 m b 0.17 m
TY 9.3.2 a 0.16 m b 2.0 kHz
TY 9.3.3 a 67.5 cm b 0.900 m
1 0.8 m
2
1.5 m
3 wavelength is 14 of fundamental wavelength
5 a 300 Hz b 600 Hz c 900 Hz
6 a 0.50 m b 0.33 m c 960 Hz
7 a 113 Hz b 563 Hz
4
0.74 m
Chapter 9 review
1
3
4
5
9
11
12
13
15
17
18
19
20
B
2
C
B, because it has higher frequency
B, because it has greater amplitude
B and C
6
A
7
C
8 D
A, B and D
10
C and D
a 2.34 kHz b 1.74 kHz
relative motion between source, observer and medium can all
cause the Doppler effect
3.1 μW m−2
14 a 100 Hz b 300 Hz
0.50 Hz.
16
9.1 cm
a 125 Hz b 2.56 m c 375 Hz
The frequency of sound in strings is dependent on the mass of the
string. Guitar strings change mass as they become thicker, so each
string can produce a different frequency.
20 mm
Responses will vary. The behaviour of sound waves can be
predicted using a wave model. The wave behaviours in this activity
are diffraction and superposition. As the waves leave the speaker
they diffract, forming a circle of sound waves around each speaker.
In areas of loud noise, the waves from each speaker arrive in
phase and constructively interfere. In areas of quiet noise, the
waves from each speaker arrive out of phase – a compression from
one wave arrives with a rarefaction from the other. The sound does
not completely cancel due to reflections and other small variations.
This interference pattern is caused by the wave nature of sound.
Chapter 10 Ray model of light
b 10 cm behind mirror
TY 10.3.3 M = − 13, h = 4 cm
TY 10.3.4 a
10.1 Light as a ray
TY 10.1.1 0.4 cd
1
TY 10.1.2 25 times brighter
observer’s eye
object
ray 2
ray 1
mirror
F
2
3
4
5
a upright b same size as object c virtual
a 5.5 m b 3.0 m s–1
When you read a book, light from your face is reflected off the
page. However, since the page is rough, the reflection is diffuse—
the light rays travel in random directions and do not form an
image.
D
6
13.5 cd
10.2 Refraction
TY 10.2.1 1.52
TY 10.2.2 1.62 × 108 m s−1
TY 10.2.3 28.2°
TY 10.2.4 24.4°
1 slower than
2
2.17 × 108 m s−1
3 1.31
4
35.3°
5 a no b yes c yes d no
image is virtual, upright and diminished
b virtual image, appears 6 cm behind lens
c M = −0.4
1 a The radius of curvature describes the distance between the
centre of a spherical mirror and the surface of the mirror.
b 25 cm
2 concave mirror converges light rays; convex mirror diverges
light rays
3 a virtual b upright c diminished
4 40 cm
5 a virtual and upright b 4.3 mm high.
6 20 cm
7
D
Chapter 10 review
10.3 Curved mirrors and lenses
TY 10.3.1 a
1
2
optical
axis
3
virtual
image
F
object
4
5
7
8
9
11
12
13
14
15
17
image is virtual, upright and magnified
b 24 mm behind mirror
TY 10.3.2 a
optical axis
18
19
ray 1
object
principal axis
image
ray 2
image is virtual, upright and diminished
v = 6 cm
F
20
21
22
23
24
25
Angle of incidence equals angle of reflection, i.e. i = r.
A real image occurs where light rays actually cross and an image
can be formed on a screen. A virtual image occurs where light rays
appear to cross; it can be seen by looking into a mirror or lens.
a virtual, upright and same size as person
b 1.5 m
a 0.75 m b 1.55 m c 0.80 m
increase by factor of four
6 2.2 cd
increases; away from
a incident ray b normal c reflected ray
d boundary between media e refracted ray
25.6°
10
2.10 × 108 m s−1
a: 25.4°, b: 25.4°, c: 28.9°
a 32.0° b 53.7° c 21.7° d 1.97 × 108 m s−1
a 19.5° b 19.1° c 0.4° d 1.96 × 108 m s−1
a 49.8° b 40.5° c 27.6°
B, D, A, C
16
1
a 80 cm
b inside focal length of mirror, i.e. closer than 40 cm to mirror
a virtual and upright, 2.7 m behind mirror
b M = 0.091
a 10 cm in front of mirror b no image formed
c 3.3 cm behind mirror
because it causes rays to diverge
a 60 cm from lens b 10 cm
a 30 cm behind lens b M = –3 c 15 mm
a virtual b 5 cm c 7.5 cm
a convex b 3.9 cm
Responses will vary. Light can be described using many models.
The ray model can predict behaviour of light during reflection
and refraction. Light travels in a straight line through a medium of
constant density. When a laser is shone through air the light travels
in a straight line. When light goes from the air into the glass, it
bends because of the change in optical density (refractive index).
ANSWERS
431
It bends again when it enters the water, and at every other change
in optical density. The water beads have nearly the same optical
density as water. When the light is travelling through the water
beads in water, it does not bend as there is no change in optical
density. When the water beads are in air there is a change in
optical density, causing the light to bend entering and exiting each
water bead.
Chapter 11 Thermodynamics
2
4
5
6
Chapter 11 review
11.1 Heat and temperature
1
2
3
TY 11.1.1 2770 J
1 C
2
a C and D b B
3 C and D
4 The temperature of the gas is just above absolute zero so the
particles have very little energy.
5 a 303 K b 102°C
6 absolute zero, 10 K, −180°C, 100 K, freezing point of water
7 −70 kJ
4
5
7
8
9
13
11.2 Specific heat capacity
TY 11.2.1 6.3 MJ
TY 11.2.2 2.1
1 water
2
aluminium
4 25.2 kJ
5
2x J
6 4.67 times that of the water
14
3
7
2100 J
B
15
11.3 Latent heat
TY 11.3.1 1.38 × 102 kJ
TY 11.3.2 7.96 × 106 J
1 a The mercury melting; temperature does not change during
phase transitions as average kinetic energy does not change.
b −39°C c 357°C d 1.26 × 104 J kg−1 e 2.85 × 105 J kg−1
2 2.25 × 105 J
3
D
4
6.7 × 104 J
16
19
21
11.4 Conduction
TY 11.4.1 500 W
1 The mass of the particles is relatively large and the vibrational
velocities are fairly low.
2 Metals conduct heat by free-moving electrons as well as by
molecular collisions. Wood does not have any free-moving
electrons.
3 thickness, surface area, nature of the material, temperature
difference between it and second material
4 Copper is a better conductor of heat
5 C
6 escaping; low; not able
7 Plastic and rubber have low conductivity, so do not allow heat
transfer from your hand very easily. Metal has high conductivity, so
heat transfers from your hand easily.
22
23
24
11.5 Convection
1
2
3
4
5
6
liquids and gases
upwards
Air over certain places, such as roads, heats up and becomes less
dense. It forms a column of rising air called a thermal.
D
It is not possible for solids to transmit heat by convection because
solids do not contain the free molecules required for convection
currents.
The source of heat, the Sun, is above the water. It takes much
longer to heat a liquid when the source is at the top, as the
convection currents also remain near the top. The warm water
is less dense than the cool water and does not allow convection
currents to form throughout the water.
11.6 Radiation
1
a partially reflected, partially transmitted and partially absorbed
b absorption
432
ANSWERS
higher; shorter; infrared
3
E
Conduction and convection require the presence of particles to
transfer heat.
a matt black beaker b glossy white surface
dark coloured metals that radiate heat energy strongly and keep
the computer cool
25
26
A
temperature
Heat refers to energy that is transferred between objects,
temperature is a measure of average kinetic energy of particles in
a substance.
a 278 K b −73°C
225 J
6
−550 J
average kinetic energy of particles in tank B is greater
balls are at same temperature
10.0 kJ
10
30°C
11
1 kg
12
B
Describes a change of state (melting). Heat energy increases the
potential energy of the particles in the solid instead of increasing
their kinetic energy.
Steam. Both have same kinetic energy as temperatures are the
same, but steam has more potential energy due to its change
in state.
The higher-energy particles are escaping, leaving behind the lowerenergy particles. The average kinetic energy of the remaining
particles decreases, thus the temperature drops.
126 J kg−1 K−1
17 7.0 kJ
18 60°C
34 kJ
20
B
Polystyrene is a good insulator, whereas metal is a good heat
conductor. Not much heat energy flows from the polystyrene to
the ice, so the ice-cube melts very slowly. Heat energy flows very
quickly from the metal into the ice, so ice-cube melts rapidly. For
the same reason, heat energy flows rapidly from your skin when
you touch the metal, causing your fingers to feel cold. Not much
heat energy flows from your fingers when you touch polystyrene,
so they do not feel as cold.
A lot of air is trapped in the down. As air is a poor heat conductor,
the quilt limits the heat transfer away from the person.
Sweat is a thin layer of water and as it evaporates, it cools down.
This is because the higher energy water molecules escape into the
air, leaving the lower energy ones behind. The sweat on your body
has a large surface area, increasing the rate of evaporation. The
breeze also increases the rate of evaporation. The cooler layer of
sweat remaining on your skin draws heat energy from your skin
and so your body cools down.
The stopper reduces heat loss by convection and conduction.
The air between the walls reduces heat transfer by conduction.
The space between the walls is almost a vacuum, so convection
currents will not form and heat will be transmitted very slowly, if
at all, between the walls. The flask’s shiny surface reduces heat
transfer by radiation.
The person emits stronger infra-red radiation than their
surroundings. The infra-red radiation is detected by the thermal
imaging technology, which allows them to be ‘seen’. The human
eye cannot always distinguish a person from their surroundings,
especially if they are under cover or if their clothes blend with their
surroundings.
Water is a better conductor of heat than air. When a person is in
cold water, the rate of energy flowing from their body is far greater
(25 times) than in cold air. In cold water, heat energy quickly
flows from the warm body decreasing the body’s temperature to
dangerous levels. A wetsuit provides a layer of insulating material
(neoprene) around the body and slows the rate of heat loss. This
prevents the person from getting hypothermia.
27 Responses will vary. When thermal energy, or heat, is added
to a closed system there are many different things that can
happen. The activity looks at how thermal energy increases the
temperature. As the thermal energy (or shaking) adds energy to
the system, the molecules in that system travel faster, increasing
the average kinetic energy. Temperature is a measure of the
average kinetic energy of the molecules, meaning that adding
heat to a system can increase the temperature. The conduction of
heat can also be considered using this model, with the heat being
transferred through a series of collisions along the material. Heat
transfer through convection can also be related to this model,
understanding that increasing the temperature increases the
space between the molecules in a fluid as the molecules move
faster. This change in density causes natural convection to occur.
32
33
34
Module 3 review
1
6
10
13
18
21
22
23
24
25
26
27
C
2 B
3
B
4
C
5
C
B and C
7
A
8
A
9
D
C and D
11
A, B, C and D
12 A and C
B
14 C
15
C
16 D
17 C
B
19
C
20
B
Temperature is related to the average kinetic energy of the
particles. On sublimation, average kinetic energy of the particles
is not altered. Potential energy increases as the molecules move
further apart.
aluminium
When heat is absorbed by a material and no phase change is
involved, the heat capacity is the energy in joules to heat 1 kg
of material by 1°C. For phase changes (latent heat) there is no
temperature change; it is merely the energy per kilogram required
to cause the phase change.
amplitude = 2 (unknown units), period = 0.004 s,
frequency = 250 Hz
A mirror creates a virtual image which can be projected onto a
screen, on same side of mirror as object. If object is sufficiently
distant, could place a piece of card at focal length of mirror and
image will be projected onto it.
557 Hz
either 443 Hz or 437 Hz; Justine needs to adjust the tension in
her string until beat disappears (then both violins will be tuned to
440 Hz)
28
35
c Responses will depend on assumptions. For this calculation,
assumed: temperature next to inner tent wall 7°C; surface area
of tent 6.0 m2, material 1.0 mm thick, thermal conductivity of
material (nylon fabric) 0.13 W m−1 K−1, no heat lost through base
of tent. In 5 hours, 39 kWh of energy lost.
a 23.0°
b 69.7°
c 46.7°
d 1.25 × 108 m s−1
a 47.5 kJ
b Water has very high specific heat capacity relative to fats and
proteins in ice cream. Ice cream mix is 70.0% water, hence its
lower specific heat.
c The heat lost to the brine at 0°C comes from latent heat of
fusion; the water changes phase and becomes ice.
d 117 kJ e −1.6°C
a 567 Hz
b 575 Hz; very small increase is unlikely to be noticed
c As an object is moving closer while emitting a sound, the
frequency of the waves becomes compressed so that it has a
higher sounding pitch. As the object moves past an observer
and the distance starts to increase, the frequency is stretched
out so that the pitch decreases.
d 264 Hz or 260 Hz
a When the driving frequency is equal to the natural frequency
of an object, resonance occurs. This causes a large amount of
vibration in the system.
b If resonance isn’t taken into account, structures can break
over time as the vibrations it causes put the structure under
immense pressure.
v
c 200
; λ = 200 m
d If driving frequency is equal to natural frequency of bridge,
resonance occurs and bridge will begin to vibrate at a large
frequency.
Chapter 12 Electrostatics
12.1 Electric charge
TY 12.1.1−6.4 × 10−6 C
TY 12.1.2 3.0 × 1013 electrons
1 They will attract (oppositely charged).
2 C
3
B
4 3.1 × 1019 electrons
5
+6.7 C
6 Copper is a good conductor of electricity because its electrons
are loosely held to their respective nuclei. Electrons move freely
through the material by ‘jumping’ from one atom to the next.
Rubber is a good insulator and prevents charge leaving.
12.2 Electric fields
object
f
image
29 6.4 m
30 a 21 cm; real
b M = −0.42; diminished and inverted
31 a Apart from air temperature, heat re-radiated from the ground
affects night temperatures. Clouds absorb some of the radiated
heat and reflect it back to Earth; on clear nights, more energy is
lost by radiation.
b Formula for transfer of energy by conduction (per unit time):
Q
kA∆T
t = d
Need to make assumptions about: temperature next to the
inner tent wall, surface area of tent, thickness of material,
thermal conductivity of material, effect of ground under tent.
TY 12.2.1 5.62 ×10−4 N C−1 in the same direction as the force
TY 12.2.2 −2.16 × 10−18 J work is done on the field
1 C
2
B
3 a true b false c false d true
e true f false g false
4 1.25 × 10−2 N
5
1.39 mC
6 5.72 × 1011 m s−2
7
1200 V
8 E = Vd has units V m−1
Since V = J C−1 = kg m2 s−2 C−1,
V m−1 = (kg m2 s−2 C−1) m−1 = kg m s−2 C−1.
E = Fq has units N C−1
N = kg m s−2
N C−1 = kg m s−2 C−1
∴ V m−1 = N C−1
12.3 Coulomb’s law
TY 12.3.1 6.32 × 10−4 N repulsion
TY 12.3.2 +6.35 × 10−10 C, −6.35 × 10−10 C
TY 12.3.3 8.0 × 105 N C−1 towards the right
1 A: positive, repulsion
B: negative, attraction
C: negative, repulsion
D: positive, attraction
ANSWERS
433
2
5
7
4
1.1 × 107 N C−1
D
3
–8.21 × 10–8 N
9000 N
6
1.435 m
a double, repel b quadruple, repel
c double, attract d quadruple, repel
Chapter 12 review
1
4
8
9
11
13
15
16
17
18
20
23
24
C
2
1.9 × 1019 electrons
3
6.7 C
3.2 × 10−19 C
5
B
6 B
7
0.0225 N
Electrical potential is work done per unit charge to move a charge
from infinity to
a point in the electric field. For two points in an
electric field (E ) separated by a distance (d ) parallel to the field,
the potential difference V is the change in electrical potential
between the two points.
25 V
10
C
field, charged particle
12
4.17 × 10−18 J
2 × 10−14 N
14
20 electrons
a work done by field b no work done
c work done on field d no work done
e work done on field f work done by field
a 1.09 × 10−19 J b work done on field
a quarter and repel b quadruple and repel
c halve and attract
5.42 × 104 m s−1
19
0.045 N, repulsive force
45.8 m
21
+1.63 × 10−4 C
22
37 N
1.97 × 1013 electrons
Responses will vary. Like charges (positive and positive or negative
and n